Flashcards Physics 3B Midterm | Quizlet

Chapter 9: Fluid Mechanics

Properties of fluids:

* What is a fluid?

* Name two fluids.

A fluid is a substance that can flow.

Liquids and gases are fluids as both substances can flow.

A gas: molecules are far apart. This makes a gas compressible. Gas molecules occasionally collide with each other, or the wall of the container.

A liquid: has a well-defined surface. Molecules are about as close together as they can get. This makes a liquid incompressible. Molecules have weak bonds that keep them close together. But molecules can slide around each other, allowing the liquid to flow.

Chapter 9: Fluid Mechanics

Properties of fluids:

* What is the density of a substance of uniform composition?

* What is the density of water at 4° C?

* Its mass (M) divided by its volume (V). That is: ρ = M/V

* The density of water at 4° C is 1000 kg/m³

Chapter 9: Fluid Mechanics

Properties of fluids:

* Compare density of liquids to density of gasses.

* Compare how liquids and gases fill their containers in our atmosphere vs. space. Understand why they differ.

* Liquids have a constant density. Gases have changing density (density of air on top of a mountain vs. density of air on ground at the bottom of the mountain).

* Liquids fill their containers, taking the shape of the container. Gases also fill their containers, taking the shape of the container. However, liquids can't fill their containers in space. Gases can. This is because, for liquids, filling their containers depends on the force of gravity.

Chapter 9: Fluid Mechanics

Pressure:

* What is average pressure (P)?

* What is the unit of pressure?

The average pressure (P) is the perpendicular component of the force (F) divided by the area (A) on which the force acts. -> P = F⊥/A

* The force exerted by a fluid on a submerged object at any point on the object is perpendicular to the surface of the object.

The unit of pressure: Pascal = Pa = 1 N/m²

Chapter 9: Fluid Mechanics

Pressure: Practice Problem

* Given the mass of air particles, and the area of the surface they're exerting a force on, find the pressure.

P = F/A

1. Using their mass, find their weight: weight = mg.

* The weight of the air particles is the force exerted on the surface they're acting on. By finding their weight, we found F.

* We have the area of the surface, so we can plug the numbers in to solve for pressure.

Chapter 9: Fluid Mechanics

Pressure:

* Compare force and pressure.

Force is a vector, but pressure is a scalar.

No direction is associated with pressure, but the direction of the force associated with the pressure is perpendicular to the surface on which the force acts.

Chapter 9: Fluid Mechanics

Variation of Pressure with Depth in a Liquid:

* When a fluid is at rest, what does this mean?

If a fluid is at rest, then all parts of the fluid are in static equilibrium.

Chapter 9: Fluid Mechanics

Variation of Pressure with Depth in a Liquid:

* Consider a liquid in static equilibrium (PIC): How is the pressure P1 related to the pressure P2?

- at different depths

- at the same depth

Consider a sample of liquid of cross-sectional area (A) and height (h) in static equilibrium. Then, the net ↑ force = 0.

----> P(deeper level) = P(upper level) + ρgh

The pressure increases as you go deeper in the liquid if it is in static equilibrium. Pressure is constant at the same depth if the liquid is in static equilibrium.

Chapter 9: Fluid Mechanics

Variation of Pressure with Depth in a Liquid:

* At sea level, what is the atmospheric pressure Po?

* How can you tell when you have atmospheric pressure in a problem?

At sea level, the atmospheric pressure Po is equal to 1.013 x 10⁵ Pa = 101 kPa = 14.7 psi = 1 atm

If you have an open container, the pressure on the fluid in the container from the air is the atmospheric pressure (Po).

Chapter 9: Fluid Mechanics

Variation of Pressure with Depth in a Liquid:

* Describe the density of air as it goes from Earth's surface -> space.

* Describe the pressure of air as it goes from Earth's surface -> space.

1. The air's density and pressure are greatest at the Earth's surface.

2. Because of gravity, the density and pressure decrease with increasing height.

3. The density and pressure approach zero in outer space.

Chapter 9: Fluid Mechanics

Variation of Pressure with Depth in a Liquid:

* Define Gauge Pressure.

* Compare Gauge Pressure and Absolute Pressure.

* Know how to calculate absolute pressure from gauge pressure.

* Should you use absolute pressure or gauge pressure in equations?

The difference between absolute pressure and atmospheric pressure is gauge pressure.

Gauge pressure = P - Po

Gauge pressure = Absolute pressure - Atmospheric pressure

Gauge pressure = P - 1 atm

Gauge pressure reads 1 atm lower than it should.

EX: If the absolute pressure = 1 atm...

gauge pressure will read: 1 atm - 1 atm = 0 atm.

Add an extra 1 atm to get absolute pressure from gauge pressure.

Always use absolute pressure to be safe, in equations that don't have pressure on both sides, the pressures won't cancel out and you will get the wrong answer.

Chapter 9: Fluid Mechanics

Variation of Pressure with Depth in a Liquid:

* Know how to calculate the pressure of liquids IN STATIC EQUILIBRIUM at different depths, when dealing with MULTIPLE DIFFERENT LIQUIDS.

* Just add more ρgh terms to represent the different liquids in the container.

Chapter 9: Fluid Mechanics

Variation of Pressure with Depth in a Liquid: PRACTICE PROBLEMS

* Two open cylinders have water in them. One has a radius of 10 m and the other has a radius of 20 m. At a depth of 50 cm from the surface, which cylinder has a higher pressure?

* Who's feet will hurt more: someone wearing heels or someone wearing regular shoes?

Since the equations for pressure don't depend on the area of the container (only the depth), both will have equal pressure.

The force however, will be different since P = F⊥/A. The one with the larger area will experience a larger force at that level. Force and area change accordingly since pressure is constant for this example.

Contrast this with a different problem. If a person wears regular shoes and later wears heels, their feet hurt more when they wear heels. The person's weight doesn't change (F⊥ is constant), so the difference in area (heels have a small area) causes the pressure to be LARGER when they wear heels. Pressure and area change accordingly since force is constant for this example.

Chapter 9: Fluid Mechanics

Pascal's Principle:

* Define Pascal's Principle.

* Describe what is happening in a hydraulic press: When a small force is applied to a small piston, describe the force applied to a piston of larger area at the same height AKA if a small force F1 is applied to the left end, describe the force F2 applied to the right end when A2 >>>>> A1.

A change in pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and to the walls of the container.

Hydraulic press: a small force is applied to a small piston. Because the pressure (P) is the same at all points at a given height in the fluid, a piston of larger area at the same height experiences a larger force.

-> A small force F1 applied to the left end results in a large force F2 applied to the right end if A2 >>>>> A1.

* F2 = (A2/A1)F1

Chapter 9: Fluid Mechanics

Pressure Measurements:

1. The open-tube manometer:

* Know what this instrument is used for.

* Know the equation for calculating pressure at different depths.

* Describe the pressure at two different points at the same depth.

This apparatus is used to measure the pressure in an enclosed fluid. The governing equation is P = Po + ρgh, where (h) represents the vertical separation distance between the levels of the liquid in the two columns of the U-shaped tube.

* The pressure is the same at the bottoms of the two tubes.

Chapter 9: Fluid Mechanics

Pressure Measurements:

* When can you use Patm for Po?

You can only use Patm for Po when you're referring to the pressure of the atmosphere, (the pressure of the air above the ocean). DO NOT use Patm when you're referring to simply the pressure of air - this doesn't work when referring to the pressure of air in a tube.

Chapter 9: Fluid Mechanics

Pressure Measurements:

2. The Barometer:

* Know what this instrument is used for, and know how this works.

* Know relevant equations.

* When there is a vacuum in a certain area, what is the pressure in this area?

This apparatus is used to measure atmospheric pressure.

* There is a near-vacuum at the top of the tube (upper level) -> P(upper) = 0 atm

* The height to which the mercury rises depends on the atmospheric pressure exerted on the mercury in the dish.

* P(deeper) = P(upper) + ρgh -> P(deeper) = ρgh

* EX: For mercury, ρ = 13.6 x 10³ kg/m³, and atmospheric pressure at sea level is Po = 1.013 x 10⁵ Pa which corresponds to height (h) of 76 cm = 0.76 m = 760 mm = 29.92 inches of mercury.

*** Using this instrument, we can measure atmospheric pressure because it always makes the pressure of the upper level zero, due to acting like a vacuum at the upper level of the tube ***

EX: in problems using this instrument, P(upper) = 0, and P(deeper) = P(atm) = ρgh

Chapter 9: Fluid Mechanics

Buoyancy Forces and Archimedes's Principle:

* Know Archimedes's Principle.

- What is F(B) (not the equation, just what it is)?

- Know the direction of the net force, and why this is its direction.

Archimedes's Principle: Any object completely or partially submerged in a fluid is buoyed upward by a force whose magnitude is equal to the weight of the fluid displaced by the object.

F(B) is the magnitude of the buoyant force acting on the submerged object, exerted by the entire surrounding fluid.

- The net force of the fluid on the submerged object is the buoyant force F(B).

F(up) > F(down) because the pressure is greater at the bottom of the object. Hence, the fluid exerts a net upward force.

Chapter 9: Fluid Mechanics

Buoyancy Forces and Archimedes's Principle:

* What is the equation(s) for F(B)?

F(B) = ρ(fluid)V(submerged)g

F(B) = ρ(fluid)V(fluid displaced)g

F(B) = W(water displaced) -> F(B) = M(water disp.)g

* M = ρV

Chapter 9: Fluid Mechanics

Buoyancy Forces and Archimedes's Principle:

* When will an object sink?

* When will an object float?

* When will an object have neutral buoyancy?

Object sinks: F(B) < Wo, ρ(avg,object) > ρ(fluid)

Object floats: F(B) > Wo, ρ(avg,object) < ρ(fluid)

Object has neutral buoyancy: F(B) = Wo, ρ(avg,object) = ρ(fluid)

Chapter 9: Fluid Mechanics

Fluids in Motion:

A fluid is flowing in a tube - how is the pressure related to the motion of the fluid? Know which equations to use.

A fluid is flowing in a tube (moving) - how is the pressure related to the motion of the fluid?

* Two equations for fluids in motion: Equation of continuity and Bernoulli's equation.

Chapter 9: Fluid Mechanics

Fluids in Motion:

* Define laminar flow.

* Know the continuity equation. Know what it results from.

* Consider an incompressible fluid (liquid) flowing through a tube: if the diameter of the tube changes, then what happens to the speed of the fluid?

- Compare the volume of the A1 cylinder to the A2 cylinder, and know how to calculate their volumes.

- Compare the mass of liquid entering vs. leaving the tube.

* What does this tell you?

- Know the limitations of using the continuity equation.

Laminar flow: when each particle of the fluid follows a smooth path so that the paths of different particles never cross each other.

The continuity equation: A1v1 = A2v2, or Av = constant

The continuity equation results from conservation of mass in laminar flow.

* The two cylinders of liquid have the same volume:

∆V1 = ∆V2

* Volume ∆V1 = A1∆x1

* Volume ∆V2 = A2∆x2

If a fluid of mass ∆M1 enters the tube through A1 during a time interval ∆t, then an equal mass of fluid ∆M2 must leave the tube through A2 (conservation of mass in laminar flow).

- Mass of fluid entering = mass of fluid leaving

- ∆M1 = ∆M2

* Equation of continuity: the amount of liquid entering the tube = the amount of liquid leaving the tube; no holes are in the tube, there's no leaking (no fluid being lost).

☆☆☆ Only used in the case of liquids ☆☆☆

- Av = constant = volume/time

Chapter 9: Fluid Mechanics

Fluids in Motion:

* Know the equation for the volume flow rate.

* Know the units for the volume flow rate.

Volume flow rate (Q): Q = ∆V/∆t = Av

* Av has units of volume/time.

Chapter 9: Fluid Mechanics

Fluids in Motion:

Volume flow rate problem:

Let's say our lecture hall is flooding with water - there is water coming in from the back doors, and a door at the front is open.

- Door at the front (open): 2 l x 2 w

- Water is moving out of the open door at 3 m/s

Calculate the volume flow rate. How much water is leaving out the front door, every second?

How much water is leaving out the front door in 3 seconds?

Let's say our lecture hall is flooding with water - there is water coming in from the back doors, and a door at the front is open.

- Door at the front (open): 2 l x 2 w = 4 m² = A

- Water is moving out of the open door at 3 m/s

- Av = (4 m²)(3 m/s) = 12 m³/s

Meaning: every second, a volume of 12 m³ goes out the door, leaving the room.

In 3 seconds, 36 m³ goes out the door, leaving the room.

Now, let's say that outside the opened door, there's a swimming pool.

* Swimming pool: 1200 m³

How long would it take to fill the swimming pool with water? Remember - 12 m³ water is filling every second.

Chapter 9: Fluid Mechanics

Fluids in Motion:

If the diameter of the tube changes, then what happens to the speed of the fluid?

* In which of these locations is the liquid moving the fastest? A1, A2, A3? Rank them, and explain.

- An increasing speed causes the diameter to ____.

- Understand example.

* In which of these locations is there the highest pressure? Rank them, and explain.

In which of these locations is the liquid moving the fastest? A1, A2, A3?

Fastest: A2

Slower: A3

Slowest: A1

An increasing speed causes the diameter to decrease.

A1v1 = A2v2 = A3v3

EX -> Garden hose: you squish it to make water come out faster, so the water reaches farther.

In which of these locations is there the highest pressure?

Biggest P: A1

Lower P: A3

Lowest P: A2

P1: pressure at A1 (bigger)

P2: pressure at A2 (smaller)

Which pressure is larger? P1 is larger.

* Pressure is the unit of force that the liquid exerts on the area of the walls of the container.

Force is bigger at A1 -> pressure is bigger at A1.

* A1: the water is moving slower here, so more pressure is building up (bigger pressure).

* A2: the water is moving faster here, so there's less pressure on the walls of the container (lower pressure).

* A2: the pressure of the water leaving the container is bigger (hits someone harder), but the pressure of the water on the container, while it's in the container, is smaller.

- At A2, the pressure of the water is only bigger on what it hits, not on the container itself.

Chapter 9: Fluid Mechanics

Fluids in Motion:

* How does speed affect diameter?

- An increasing speed causes the diameter to ___.

An increasing speed causes the diameter to decrease.

Chapter 9: Fluid Mechanics

Fluids in Motion:

* What about a gas? What happens to the speed of a gas when A ↓?

Gases

As A ↓, the speed of the gas ↑. Quantitatively, gases don't follow: Av = constant

but ... qualitatively, they do.

Chapter 9: Fluid Mechanics

Ideal Fluid Dynamics:

Bernoulli's Equation

* What does this equation apply to?

* What does this equation result from?

* Know the equation.

- How does the equation change if h1=h2?

* Understand applications of Bernoulli's principle.

- Faster moving liquid = ___ pressure exerted by the liquid.

* Know how to calculate lift force, and know what makes it larger/smaller.

Bernoulli's Equation:

Applies to an ideal liquid with zero friction-like forces.

Results from conservation of energy. Applying the work-energy theorem to the laminar flow described below yields:

P₁ + (1/2)ρv₁² + ρgh₁ = P₂ + (1/2)ρv₂² + ρgh₂

OR

P + (1/2)ρv² + ρgh = constant

* If h1 = h2, then ... P + (1/2)ρv² = constant

Applications of Bernoulli's principle:

1. Blowing air over sheet of paper in front of your mouth.

2. Canvas top puffs upward in moving convertible cars.

3. Houses may explode during hurricanes or tornados.

4. Lift force on airplane wings:

- Lift force = (pressure difference) x (area of wing)

Lift is greater when the wing area is large or when the plane moves fast so that the pressure difference across the top and bottom of the wings is large.

As you hold a piece of paper, it bends.

* The pressure above the paper = the pressure below the paper ----------> cancel.

* The weight of the paper is the only force left.

* The faster air moves, the lower the pressure (blowing air on paper).

* While blowing air on paper, the paper lifts up, making itself flat.

A1v1 = A2v2

Chapter 9: Fluid Mechanics

Ideal Fluid Dynamics:

Magnus effect

* Understand the Magnus effect.

- Compare the motion of air relative to a non spinning ball vs. the motion of a spinning ball

- Motion of a spinning ball:

* Which side of the ball moves opposite to the airflow?

* Which side of the ball moves in the direction of the airflow?

- Describe the force generated when a spinning ball moves through air.

* A moving ball ____. So, when air moves past a spinning ball, what happens?

* In what direction does the resultant force point?

* Understand example given.

Magnus effect:

(a) motion of air relative to a nonspinning ball

(b) motion of a spinning ball

* This side (top) of the ball moves opposite to the airflow.

* This side (bottom) of the ball moves in the direction of the air flow.

(c) force generated when a spinning ball moves through air

* A moving ball drags the adjacent air with it. So, when air moves past a spinning ball:

- On one side, the ball slows the air, creating a region of high pressure.

- On the other side, the ball speeds the air, creating a region of low pressure.

The resultant force points in the direction of the low-pressure side.

Soccer field:

* A goalie kicks the ball - the ball starts spinning clockwise in the air, moving further away, the air moves faster -> less pressure is on the ball.

- Slower moving air = more pressure.

Chapter 9: Fluid Mechanics

Buoyancy Force: Liquid Under Static Equilibrium

* Know how to solve this kind of problem (QUIZ 1)

Static equilibrium -> the sum of the forces = 0.

* F(B) is one of the forces.

F(B) = ρ(fluid)V(submerged)g

F(B) = ρ(fluid)V(fluid displaced)g

F(B) = W(water displaced) -> F(B) = M(water disp.)g

* M = ρV

Chapter 9: Fluid Mechanics

Practice Problem:

1. A bargain hunter purchases a "gold" crown at a flea market. After he gets home, he hangs the crown from a scale and finds its weight to be 7.84 N. He then weighs the crown while it is completely submerged in water of density 1000 kg/m³, and now the scale reads 6.86 N. Is the crown made of pure gold?

* The reading on the scale is the measure of F the scale is putting upwards on the crown.

The tension of the cord holding the crown = the number the scale reads.

* Crown is at rest, F(net) = 0.

* T = 7.84 N

* 7.84 N - Mg = 0

* 7.84 N = Mg

* 7.84 N/g = M

M(crown) = 7.84/9.8 = 0.8 kg

The reading on the scale when the crown is submerged in water is different. When the crown is submerged in water, its reading is 6.86 N, which is the force the string is putting on the crown.

* Weight = Mg = 7.84 N

* Crown is at rest, so F(net) = 0.

* The crown should be going down, this doesn't happen because F(B) pushes it up.

T + F(B) - Mg = 0

6.86 + F(B) - 7.84 = 0

7.84 - 6.86 = F(B) = 0.98 N

AKA

F(B) = ρ(water)V(crown)g = 0.98 N

(1000)(V)(9.8) = 0.98 -> 0.001 m³

ρ(crown) = M(crown)/V(crown) = 0.8 kg/0.001 m³ = 8000 kg/m³

ρ(gold) = 19,300 kg/m³, so the crown is either hollow or it is made out of some alloy.

Chapter 9: Fluid Mechanics

Practice Problem:

2. The approximate radius of the aorta is 1.2 cm, while that of a capillary is 4.0 µm. The approximate average blood flow speed is 40 cm/s in the aorta and 0.05 cm/s in the capillaries. If all the blood in the aorta eventually flows through the capillaries, estimate the number of capillaries in the human circulatory system. Here you may assume that the blood flow rate in the aorta must equal the blood flow rate in all capillaries.

* Blood entering = blood leaving

V(blood) from A1 in 1 s = (V(blood) from A2 in 1 s)(# capillaries)

A1v1 = A2v2 x (N)

* N = # capillaries

Chapter 9: Fluid Mechanics

Bermouidilli's Principle:

* Understand how Bermouidilli's Principle relates to blood flow in human blood vessels: blood clots vs. vessel expansion

P + (1/2)ρv² + ρgh = constant

* The flow of blood is horizontal.

* Let's say someone accumulates clots in the inner walls of their blood vessel.

- When blood goes through a narrow constriction (with clot), is it going faster or slower? FASTER.

- Moving faster = pressure is less at that point.

- Moving slower = pressure is greater at that point.

- As the blood is moving through, the pressure is the force of the blood on the walls of the vessel per unit area.

- The blood vessel has no vacuum - it has stuff pushing into it -> F is being exerted on the blood vessels' outer walls -> balance out.

- With narrower walls (clot), the F pushing on the wall from the outside isn't as strong at that point.

- Clot accumulating = blocks pathway completely.

* Blood vessel expands in certain area = blood is moving slower than in a normal vessel.

- Moving more slowly = bigger pressure

- The F exerted by the slower moving blood is bigger, so that region of the blood vessel expands even more, making the problem even worse -> blood vessel could rupture.

Chapter 12: Thermodynamics

Heat

Temperature and Thermal Equilibrium

* Define temperature.

* Define thermal equilibrium: when are two systems said to be in thermal equilibrium?

Temperature and Thermal Equilibrium

* Temperature is a measure of how hot or cold something is.

* Two systems are said to be in thermal equilibrium with one another if and only if the two systems have the same temperature.

- Temperature determines whether something is in equilibrium with something else.

Chapter 12: Thermodynamics

Heat

The Zeroth Law of Thermodynamics

* What is the Zeroth Law of Thermodynamics? (what does it state):

If objects A and B are separately in thermal equilibrium with a third object C, then objects A and B are ___.

* What are thermometers: what do they measure, and how?

* Know the difference between T(F), T(C) and T.

- Which is "absolute temperature"?

* Know the metric system's unit of temperature.

The Zeroth Law of Thermodynamics: if objects A and B are separately in thermal equilibrium with a third object C, then objects A and B are in thermal equilibrium with each other.

- EX: If a red marker is at the same temperature as a blue marker (thermal equilibrium), and the blue marker is in thermal equilibrium with a black marker, then the black marker is also in thermal equilibrium with the red marker.

Thermometers: instruments used to measure temperature, usually by the expansion or contraction of a liquid (such as mercury).

T(F) = Temperature in degrees Fahrenheit

T(C) = Temperature in degrees Celsius

T = "Absolute temperature" in Kelvin

- Metric system: unit of T = K.

↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑

You'll need to make sure temperature is always converted to K (or equivalent) when using formulas.

Chapter 12: Thermodynamics

Heat

The Absolute Zero Temperature

* Define the "absolute zero temperature".

* Define the "zero-point energy".

* At what temperature does water freeze at 1 atm? Boil?

* Compare the Celsius and Kelvin scale.

* When using temperatures in equations, what is a good rule of thumb to follow?

* Define ∆T.

The "absolute zero temperature" is the temperature at which the system is in a state of minimum but finite atomic motion. The "zero-point energy" is the energy associated with this state of the system, i.e. with the motion of atoms at 0 kelvin.

☆☆☆ MEMORIZE WHAT TEMPERATURE WATER FREEZES AND BOILS AT, AT 1 ATM ☆☆☆

- It's easiest to memorize on the Celsius scale

-> (Boils @ 100, Freezes @ 0).

- Note: values in table are only true when P = 1 atm.

The change in temperature of the Celsius scale is the same as the change in temperature of the Kelvin scale, regardless of the position of the temperature.

---> ∆T(C) = ∆T(K)

* TIP: Always convert temperature to Kelvin when given a temperature in a problem.

If ∆T is in the problem, you can just use Celsius, if C is given, since ∆T(C) = ∆T(K). If no ∆T is in the problem, then convert the temperature to Kelvin.

∆T = T(hot) - T(cold)

*** T⁴ isn't ∆T⁴, just T! -> You must convert T to Kelvin -> If you just have T by itself, convert to Kelvin.

T⁴ - Ts⁴: must convert to Kelvin.

Chapter 12: Thermodynamics

Heat

The Three Phases of Matter

1. A solid is a ___ system that consists of ___ atoms connected by ___ molecular bonds. Solids are nearly ___, which tells us that the atoms in a solid are ___.

- Atoms ___.

2. A liquid is a system in which the molecules are ___ held together by ___ molecular bonds. The bonds are ___.

- Atoms are held ___ by ___ molecular bonds, but they can ___.

3. A gas is a system in which ___ until, on occasion, it ___. A gas is ___, telling us that there is ___.

- Atoms are ___ and ___ except for ____.

1. A solid is a rigid macroscopic system that consists of particle-like atoms connected by spring-like molecular bonds. Solids are nearly incompressible, which tells us that the atoms in a solid are just about as close together as they can get.

* Atoms vibrate around equilibrium positions.

2. A liquid is a system in which the molecules are loosely held together by weak molecular bonds. The bonds are strong enough that the molecules never get far apart but not strong enough to prevent the molecules from sliding around each other.

* Atoms are held close together by weak molecular bonds, but they can slide around each other.

3. A gas is a system in which each molecule moves through space as a free particle until, on occasion, it collides with another molecule or with the wall of the container. A gas is compressible, telling us that there is lots of space between the molecules.

* Atoms are far apart and travel freely through space except for occasional collisions.

Chapter 12: Thermodynamics

Heat

Heat & Heat Units

* Define Heat: the ___ energy being transferred, via ___, from a region of ___ to a region of ___.

* Define 1 calorie.

* 1 calorie = how many Joules?

* 1 diet Calorie = how many calories?

* How many Joules are required to raise the temperature of 1 kg of water by 1 °C?

Heat is the thermal energy being transferred, via particle collisions, from a region of high temperature to a region of lower temperature.

* Energy in transit from point to point (usually hot to cold).

* 1 calorie is defined as the energy needed to raise the temperature of 1 gram of water by 1 °C.

* 4186 Joules are required to raise the temperature of 1 kg of water by 1 °C.

- 1 calorie = 4.186 Joules

- 1 diet Calorie = 1000 calories

Chapter 12: Thermodynamics

Heat

Heat & Heat Units: Examples

* Let's say we have a metal rod with one end in fire and the other end in ice (one end is hot, one end is cold).

- Describe the atoms at the hot end of the rod.

- What starts happening to these atoms as the temperature continues to rise? What happens to the rod?

- Summarize what is happening here.

- Define heat energy.

* Understand example given.

EX (Conduction) -

Let's say we have a metal rod, with one end in fire and the other end in ice -> one end is hot, one end is cold.

* Hot end of rod: atoms aren't moving around, they're just dancing in the same space given to them (think of a crowded lecture hall, with students shoulder to shoulder in their seats. students are moving during the lecture, taking notes, but not leaving their seats). Molecules in the solid are in equilibrium positions, but vibrating in these equilibrium positions.

* Atoms start jiggling more rigorously with rising temperatures -> solid expands. This jiggling motion of agitation of these atoms leads to them agitating the atoms next to them -> domino effect.

SUMMARY: The fire on one end of the rod -> transfers energy of agitation from hot to cold -> this energy is heat energy.

EX: The Sun is hot - there's a vacuum between our Sun and the Earth. Heat isn't transferred to the Earth from convection nor conduction - it is transmitted by waves.

Chapter 12: Thermodynamics

Heat

Thermal Expansion of Solids and Liquids - Linear Expansion of Solids

* Know the equation for linear expansion OF SOLIDS.

- Know when this equation can be used.

The length L₀ of a solid object (or thin liquid) changes by an amount ΔL when its temperature changes by an amount ΔT given by: ΔL = L₀αΔT (solids)

where...

α = thermal coefficient of linear expansion of the object.

For many materials, every linear dimension changes according to the formula above. Thus, L could represent the thickness of a rod, the side length of a square sheet, or the diameter of a hole.

* EX: ΔL = change in length of wire.

Chapter 12: Thermodynamics

Heat

Thermal Expansion of Solids and Liquids - Volume Expansion of Liquids and Solids

* Know the equation for volume expansion OF LIQUIDS AND SOLIDS

- How does β compare to α?

The volume V₀ of a solid object, or liquid, changes by an amount ΔV when its temperature changes by an amount ΔT given by: ΔV = V₀βΔT (liquids and solids)

where...

β = thermal coefficient of volume expansion of the solid or liquid

* Water ----heated----> grows in space = change in volume.

* β = 3x α

Chapter 12: Thermodynamics

Heat

Thermal Expansion

* Describe thermal expansion.

* Understand the idea of linear expansion.

* Let's say we have two markers - one marker is taller than the other:

- Two markers: some length, some temperature.

- Both of the markers have the same initial temperature, and we raise their temperatures by the same amount.

Will ΔL be the same, or different?

- SUMMARY: If our green marker is longer than our red marker, if the red marker increases by some amount, the green will ___ by a ___ amount. How much it ____ in length depends on ___.

- Which marker (green or red) undergoes the greater fractional change in length?

Thermal expansion: if you add heat energy -> atoms jiggle more vigorously -> atoms spread out -> grows in volume or length.

* α = different values of linear expansion for different elements.

* Let's say you have a rod - initial length (L₀), temp (T).

- You add heat to it -> rod expands.

- If you raise the temperature by 10 °C/K, of a rod with L₀, the rod gets longer. How much longer?

-> We've just experienced a change in the length of the rod as a result of the change in T.

* Let's say we have two markers - one marker is taller than the other:

- Two markers: some length, some temperature.

- Both of the markers have the same initial temperature, and we raise their temperatures by the same amount.

Will ΔL be the same, or different?

ΔL = L₀αΔT

L₀ is different between the markers -> the change in length of the bigger marker will be bigger.

* α is the same for both markers (same material), and both have the same ΔT.

* SUMMARY: if our green marker is longer than our red marker, if the red marker increases by some amount, the green will increase by a bigger amount. How much it increases in length, tho, depends on how long it was to start with.

EX: "Which marker (green or red) undergoes the greater fractional change in length?"

SAME: fractional change = ΔL/L₀ = αΔT

Chapter 12: Thermodynamics

Heat

Thermal Expansion of Solids and Liquids - Holes

* What happens to a hole in a piece of material when heated and cooled?

* When an object undergoes thermal expansion, what happens to any holes in the object?

- EX: Let's say you heat up your ring - does the hole get smaller or bigger?

* Understand example given.

A hole in a piece of material expands when heated and contracts when cooled, just as if it were filled with the material that surrounds it.

* When an object undergoes thermal expansion, any holes in the object expand as well (the expansion is exaggerated).

- A plate expands when heated -> so a hole cut out of the plate must expand, too.

Chapter 12: Thermodynamics

Heat

Thermal Expansion of Solids and Liquids - Examples

* Expansion slots on bridges are needed to accommodate changes in ___ that result from ___ throughout the seasons.

* Running hot water on a jar lid -> ____. This is because the metal may be heated ___ by the hot water, or even if the jar is heated uniformly, metals ___ than glasses because ___.

* Understand examples.

Expansion slots on bridges are needed to accommodate changes in length that result from thermal expansions and contractions throughout the seasons.

Running hot water on a jar lid loosens it. This is because the metal may be heated more directly by the hot water, or even if the jar is heated uniformly, metals expand more than glasses because

α(metal) > α(glass)

* Golden Gate Bridge: increases each summer by a meter.

* What happens if you pour a sidewalk in the winter time? What happens when the summer comes along? Slabs on concrete expand. If you don't allow for expansion, they can bunker.

* Bridges expand in sections, so they have expansion spots. When the slabs grow, they grow fitting like a puzzle, with each other, and in the winter, they move apart, leaving space between (PIC).

* Jar: If you're having a hard time opening a jar, run it under hot water -> expands because metal has a bigger alpha coefficient than glass, so you can open it.

Chapter 12: Thermodynamics

Heat

Thermal Expansion of Solids and Liquids - Water

* Water ___ as its temperature increases from 0°C to 4°C, and thus its ___ in this temperature range.

* Notice that above 4°C, water exhibits ___ with increasing temperature.

The anomalous behavior of water between 0°C and 4°C:

Water contracts as its temperature increases from 0°C to 4°C, and thus its density increases in this temperature range.

It is this weird behavior of water in this temperature range that explains why lakes freeze from the top down.

Notice that above 4°C, water exhibits the "expected" expansion with increasing temperature.

Chapter 12: Thermodynamics

Heat

Specific Heat (Solids and Liquids)

* Know the equation for heat energy under ONE PHASE.

- Know what calculating this (Q) value tells you.

- What are its limitations?

- Know the variables of the equation, and what they stand for.

- What is the specific heat (capacity) of water?

The heat energy (Q) that must be supplied (or removed) to change the temperature of a solid (or liquid) substance of mass (m) by an amount ΔT as it remains in the same phase, and at constant pressure, is given by:

Q = mcΔT (solids and liquids, one phase only)

where...

c = the specific heat (capacity) of the substance.

Water has a relatively high value of specific heat,

c(water) = 4186 J/kg °C

Chapter 12: Thermodynamics

Heat

Molar Specific Heats (gases)

* Know the equation for heat energy, involving molar specific heat.

- When is this equation used?

* Know the variables of the equation, and what they stand for.

* What is another way to define molar specific heat?

For gases, one introduces the molar specific heat (C) as ...

Q = nCΔT

where n is the number of moles of the gas.

Also, one may define the molar specific heat C as

C = (1/n)(dQ/dT)

Chapter 12: Thermodynamics

Heat

Phase Changes & Latent Heat (L)

A. Melting & Freezing:

B. Vaporization & Condensation:

C. Latent Heat of Combustion:

* Define Latent Heat. Limitations?

A. Melting & Freezing:

* Know the equation for heat energy during this phase change.

- Know the variables of the equation, and what they stand for.

=> What is the (L) value of this phase change called? Know what this means.

* When would Q be (+) vs. (-)?

* What is the L(f) for water?

B. Vaporization & Condensation:

* Know the equation for heat energy during this phase change.

- Know the variables of the equation, and what they stand for.

=> What is the (L) value of this phase change called? Know what this means.

* When would Q be (+) vs. (-)?

* What is the L(v) for water?

C. Latent Heat of Combustion:

* What processes are analogous to phase changes? How?

What does a lower L value mean?

The Latent Heat (L) is the heat energy per kilogram that must be added or removed when a substance changes from one phase to another (at a constant temperature).

A. Melting and Freezing:

Q = ±mL(f) (phase change)

where...

m = mass of the substance undergoing the phase change.

L(f) = latent heat of fusion = energy required to break all the intermolecular bonds in one kilogram to convert the solid phase to the liquid phase.

(positive sign for melting, negative sign for freezing).

For water, L(f) = 3.33 x 10⁵ J/kg

B. Vaporization and Condensation:

Q = ±mL(v) (phase change)

where...

m = mass of the substance undergoing the phase change.

L(v) = latent heat of vaporization = the energy required to break all the intermolecular bonds in one kilogram so as to convert the liquid phase to the gas phase.

(positive sign for vaporization, negative sign for condensation).

For water, L(v) = 2.26 x 10⁶ J/kg = 22.6 x 10⁵ J/kg

C. Latent Heat of Combustion:

Chemical reactions such as combustion are analogous to phase changes in that they involve definite quantities of heat.

EX: Complete combustion of 1 gram of gasoline produces about 46,000 J or about 11,000 calories. Thus, the latent heat of combustion L(c) of gasoline is:

L(c) = 4.6 x 10⁷ J/kg (combustion of gasoline)

which is about 46 million Joules per kilogram of gasoline.

Energy values of foods are defined similarly.

EX: When we say that a gram of peanut butter "contains 6 calories", we mean that 6 cal of heat (6000 cal or 25,000 J) is released when the carbon and hydrogen atoms in the peanut butter react with oxygen and are completely converted to CO2 and H2O.

* A lower L value = easier to melt/boil/freeze (requires less energy).

Chapter 12: Thermodynamics

Heat

Calorimetry:

* Know the rules for calorimetry:

- When using equations of the type "heat gained = heat lost", what rules must we follow? Therefore, when calculating heat energy contributions, how should temperature changes be measured?

Phase Diagrams: SOLID -> LIQUID -> GAS

* In a phase diagram, what is occurring at the steps where Q = mcΔT would be used?

* In a phase diagram, what is occurring at the steps where Q = +mL would be used?

Equations of the type "heat gained = heat lost"

* Here, both sides of the equation must have the same algebraic sign. Therefore, when calculating heat energy contributions, always write any temperature changes as the higher temperature minus the lower temperature!

* Temperature of water changes: during these periods, temperature rises as heat is added -> Q = mcΔT

⋆˚✿˖° Ice warms

⋆˚✿˖° Liquid water warms

⋆˚✿˖° Steam warms

* Phase of water changes: during these periods, temperature stays constant and the phase change proceeds as heat is added: Q = +mL

⋆˚✿˖° Ice melts to liquid water at 0°C

⋆˚✿˖° Liquid water vaporizes to steam at 100°C

Chapter 12: Thermodynamics

Heat

Q = mcΔT VS. Q = mL

* What is the broad definition of Q (applying to both equations)?

* Know the equations for Q.

* Know the conditions that must be satisfied in order to use each Q equation.

* If a system undergoes ΔT, it cannot undergo ___.

Q = the amount of heat energy being transferred (in CH. 12)

1. Q = mcΔT

2. Q = mL

1. Conditions that need to be satisfied to use Q = mcΔT:

* Good for solids and liquids (not gases)

- If a solid, m = mass of the solid.

- If a liquid, m = mass of the liquid.

* Q (in this equation) = the heat energy needed to be added or subtracted from the solid of m, or liquid of m, such that they undergo a temperature change of ΔT.

- c = a specific value of a certain liquid or solid.

* If a system undergoes ΔT (temperature change), it cannot undergo a phase change.

- 3 kg of a liquid -> 3 kg of a liquid.

Doesn't boil or freeze, remains in the same phase.

* ONE PHASE ONLY, NO PHASE CHANGE.

=> What happens when you use (Q = mcΔT), incorrectly:

EX: you have ice (solid), and the ice begins to melt into water, while it is melting (while phase-changing is taking place - such as condensing, steaming, etc.), the temperature of the system stays constant -> ΔT = 0 -> Q = 0.

* This would be saying that no energy is required to go from solid -> liquid (NOT TRUE).

What is our solution? Use (Q = mL) instead, for phase changes!

2. Q = mL

* Good for phase change

(solid -> liquid, liquid -> solid, liquid -> gas, gas -> liquid)

* m = the mass undergoing the phase change (might not be the entire thing, maybe only 2g out of 10g is melting).

Chapter 12: Thermodynamics

Practice Problem

2. How much heat energy must be added to 5 kg of ice initially at -10°C to transform it into 2 kg of water, and 3 kg of steam? The atmospheric pressure is 1 atm.

FOCUS: at the end, we want a combination of water and steam.

1. Ice will heat until it gets to 0 °C.

* We need to add energy to go from -10 °C -> 0 °C.

2. At 0 °C, we need to melt the ice (the entire 5 kg).

- Stays at 0 °C (phase change).

3. We need to raise the temperature of water to 100 °C (bp), all of it (5 kg).

4. We need to add more energy to the water to get 2 kg of water at 100 °C + 3 kg of steam at 100 °C.

- To do this, we need to add a certain amount of energy: Q4.

3. Final answer: Q(total) -> Q(total) = Q1 + Q2 + Q3 + Q4

Q1: the energy required to go from -10 °C -> 0 °C

* 5 kg -> 5 kg, without phase changes -> Q = mcΔT

* ΔT = 10

=> 10 °C = 10 °K -> don't need to convert.

Q2: when the ice is at 0 °C, melting to water at 0 °C

* Q = mL (phase change)

* We're converting the entire 5 kg of ice -> 5 kg water.

Q3: we're raising the temperature from 0 °C -> 100 °C

* 5 kg water -> 5 kg water

* As the water is going up from 0 °C -> 100 °C, we still have water at 100 °C (no phase change).

- Q = mcΔT

* c = the specific heat capacity of water.

Q4: when we have 5 kg of water at 100 °C, and then boil 3 kg out of 5kg to make steam, leaving 2 kg of water left.

* When we add energy, we don't want to boil all of it - we want to keep 2 kg of water.

* The temperature isn't changing, only a phase change is happening -> Q = mL

- L depends on the material undergoing the phase change.

- What "m" value do we use in this calculation? 5, 3, or 2 kg?

=> 3 kg is undergoing the phase change. We're boiling 3 kg of water to make 3 kg of steam, so m = 3 kg.

- L of Q4 = L(v)

Chapter 12: Thermodynamics

Practice Problem: BACKGROUND INFO

3. In an insulated vessel, 457 grams of ice initially at 0 °C are added to 500 grams of water initially at 15 °C .

a. Calculate the final equilibrium temperature of the system.

b. If your answer comes out to 0 °C , then what mass of ice remains?

For water, the latent heat of fusion is 3.33 x 10⁵ J/kg and the specific heat capacity is 4186 J/(kg°C).

The atmospheric pressure is 1 atm. Ignore any heat energy exchanges to the insulated vessel and the surroundings.

*** Make sure you know and understand all of this ***

* Insulated = any change in energy occurs within the vessel (no leaking in or out).

Ice is at 0 °C, water is at 15 °C

* Energy is transferred from the water to the ice (hot to cold, always). As soon as the thermal energy goes from water -> ice, the water won't change phase right away (it won't turn straight to ice right as energy is transferred), so there has to be a ΔT. This means that the final temperature cannot be 15 °C. Therefore, the limit on what the temperature could be is: it cannot be less than 0 °C (being heated up, not possible), and it cannot be greater than or equal to 15 °C (not equal because there has to be a ΔT, and not greater because we're adding ice to the water, so it has to be cooling down).

When ice receives energy -> melts (staying at 0 °C). When the water drops down in temperature from 15 °C -> 0 °C, it is possible that the ice hasn't melted completely, so the final temperature could be zero.

* As the cold ice warms up, the hot water cools down. This keeps happening until the water and ice meet at thermal equilibrium (when they both reach the same temperature).

* The total energy transfer = 0 (insulated).

* We have hot water and cold ice - the energy is leaving from the hot water and being transferred to the cold ice.

* In rigid thermal equilibrium, energy given up = energy received, or the energy lost by the hot water = the energy gained by the ice.

* Be careful with Q = mcΔT.

Make sure you have a positive # on both sides of the equation (energy given up = energy lost). Make sure energies are positive. With (higher T - lower T) we get no negative signs.

˗ˋˏ ♡ ˎˊ˗ ΔT: higher temp (whether final or initial) - lower temp.

NOT Tf-Ti

Chapter 12: Thermodynamics

Practice Problem: FINAL ANSWER

3. In an insulated vessel, 457 grams of ice initially at 0 °C are added to 500 grams of water initially at 15 °C .

a. Calculate the final equilibrium temperature of the system.

b. If your answer comes out to 0 °C , then what mass of ice remains?

For water, the latent heat of fusion is 3.33 x 10⁵ J/kg and the specific heat capacity is 4186 J/(kg°C).

The atmospheric pressure is 1 atm. Ignore any heat energy exchanges to the insulated vessel and the surroundings.

ANSWER:

Thermal energy is going from water to ice.

* If not all the ice melts, then we need to use (Q = mL).

* How much energy is needed to melt the ice?

The total energy needed to melt all the ice -> Q = mass ice x L.

-> Q = 0.457 x 333000 = 152181 J.

If the ice receives this much energy, all of the ice will melt.

The total energy given up or released by water, if and only if, it cools down to 0 °C -> Q = m(water)c(water)ΔT.

* Under conditions: water cools from 15 °C -> 0 °C

Is it going to cool down to 0 °C? Or will it be more than 0 °C?

* Energy from water -> ice.

* Water is hot at 15 °C, compared to ice at 0 °C. As water goes down in temperature, it is sending energy to ice.

* Energy needed to melt all of the ice = 152181 J needed.

BUT - water going down to 0 °C can only give up 31,425 J. There's not enough energy to melt the ice completely. So, the final temperature has to be 0 °C (since there will still be ice remaining).

Heat energy gained by ice -> Q = mL

* The mass used (m) is only the mass that WILL melt.

* Final equilibrium temperature = 0 °C.

* 0 °C, so ice remaining = 457 g (started with) - 94 g (lost) = 263 g ice remains.

Chapter 12: Thermodynamics

Heat

Mechanisms of Heat Transfer - Conduction, Convention and Radiation

A. Conduction

A. Conduction:

* This is the process whereby heat energy is transferred directly through a material (solid), with any bulk motion of the material playing no role in the energy transfer. Thermal conduction takes place via:

(i) exchange of kinetic energy between atoms (collisions)

(ii) motion of free electrons

The rate of energy transfer ("power" or "heat current")

(dQ/dt) by thermal conduction through a solid is given by:

dQ/dt = [kA(T(hot) - T(cold))]/L (in Watts)

where...

k = thermal conductivity of the material

A = cross-sectional area of the material

L = length of the material (from the hot end to the cold end)

T(hot) - T(cold) = temperature difference between the hot and cold ends of the solid.

* The thermal energy always travels from hot to cold.

* T(hot) and T(cold) are maintained by reservoirs.

- This material is conducting heat across the temperature difference.

EX: Hot rod

Q per unit of time (heat energy per second) = how much heat energy flows through the cross section area in one second.

* Energy in J/s depends on the cross section area the energy will flow through - the bigger the area that energy flows through, the more energy you'll get.

* Energy is inversely proportional to the length between hot→cold.

* The bigger the conductivity (k), the more energy will go through.

Chapter 12: Thermodynamics

Mechanisms of Heat Transfer - Conduction, Convention and Radiation

B. Convection

This is the process in which the total energy is carried from a hot region to a colder region by the bulk movement of a fluid.

There are convection currents in a pan of water being heated by a flame.

* Convection: the process by which thermal energy is transferred from hot→cold by the actual movement of a fluid (liquid or gas), flowing past a hot object and then bringing that energy to somewhere it is cold.

- PIC: a liquid is above fire, the fire is heating the water above the bottom of the container. The water will expand as it heats up → expands → density gets smaller.

* Volume increases, mass stays the same → density drops.

* Hot water rises up, cold water goes down where it'll get hot and rise up.

* Flow of liquid from hot end to cold end (convection currents generated → heating up entire body of water).

EX: Resistor

* A resistor is connected to voltage → turned on → electric current.

* Resistor gets hot, and if connected to a fan, the fan will turn. Between the fan and the resistor, there's cool air. The fan is blowing cool air through the resistor → forms hot air. Thermal energy is transferred from the hot air → cold air by the actual movement of fluid (air). Air does the job of transferring thermal energy from hot→cold.

Chapter 12: Thermodynamics

Mechanisms of Heat Transfer - Conduction, Convention and Radiation

C. Electromagnetic Radiation

This is the process whereby energy is transferred by means of electromagnetic waves. Every object emits electromagnetic radiation because of the thermal motion of its atoms or molecules on its surface.

Stefan - Boltzman Law of Radiation:

The rate at which an object emits energy by thermal radiation is:

dQ/dt = 𝑒𝓸AT⁴ (in Watts)

where ...

𝓸 = Stefan - Boltzmann constant = 5.67 x 10⁻⁸ W/m²K⁴

A = surface area of the object

T = absolute temperature on surface of the object

𝑒 = emissivity, 0 ≤ 𝑒 ≤ 1. This is equal to the fraction of the incident radiation that is absorbed by the surface.

Note from the definition of emissivity that an object that is a good absorber is also a good emitter.

Do NOT confuse emission with reflection!

Absorption and Reflection are opossite processes. A good absorber reflects very little radiant energy. Hence a surface that reflects very little or no radiant energy looks dark. Similarly, poor absorbers are also good reflectors.

EX: Clean snow is a good reflector and therefore does not melt rapidly in sunlight.

In the summer, light colored buildings stay cooler because they reflect much of the incoming radiation (good reflector thus poor absorber of radiation).

In the winter, light-colored buildings also stay warmer because they are poor emitters (since poor absorber) so they retain much of their internal energy compared to darker buildings.

=> Paint your house white if you wish to conserve energy (and save $$$)!

Electromagnetic radiation:

* Every object emits electromagnetic waves.

* How much energy is emitted from the surface of the object, where energy comes out, in 1 second depends on how hot the object is.

☆☆☆ Be careful - no ΔT (convert to K) ☆☆☆

* There is tremendice dependence on the absolute temperature of the surface of the object.

* Some objects emit more energy than others. If two objects have the same surface area and temperature, then the only difference between them is the nature of the material of their surface = the only thing different is emissivity = emit different amounts of energy.

☆☆☆ Memorize Stefan-Boltzmann constant ☆☆☆

Emissivity:

* Object #1 - solid object, not transparent - it is opaque. We have radiation incident on it, falling on it from the outside. Say 100 J are incident of this opaque object.

- Opaque = can only be reflected or absorbed, doesn't go through.

If the object reflects 5 J out of 100 J of energy shining on it, then the absorbed energy is 95 J. So ...

→ emissivity = (energy absorbed)/(incident energy)

= (95 J)/(100 J )= 95% = 0.95 (unitless quantity).

- Absorbed 95 J → good absorber.

- Only reflected 5 J → poor reflector.

* Object #2 - also opaque. There is 100 J of energy coming in, the object reflects 90 J → energy absorbed is 10 J. So...

e = (energy absorbed/incident energy) = (10/100) = 10% = 0.1

- Poor absorber (only absorbed 10%).

- Object reflected a bunch of energy incident on it = good reflector.

Energy is either reflected or absorbed.

EX: Black cap

* White light is shining on the cap.

* Black = absence of light → much is reflected from it → good absorber, poor reflector.

Good absorbers tend to be on the black side.

If we keep track of the temperature of the black cap and find that it's the same ...

* White light is shining on it - we know it's black, so it's absorbing a lot and reflecting very little. If it absorbs so much energy, the temperature should be going up and the cap should be beginning to melt. The energy going into the cap should raise the temperature until it hits the melting point → liquid heats up → reaches boiling point.

* The temperature is the same - this means that it has to be emitting as much energy as it is absorbing.

☆☆☆ Don't confuse emission with reflection ☆☆☆

* If the temp stays the same, the cap must be emitting from the surface, and emitting as much as it is absorbing. The cap does remain the same, so it is emitting as much energy as it is absorbing. It is absorbing a lot of energy because it is black, which means that it is not reflecting much (not white).

* Good absorber with temperature remaining the same = good emitter.

EX: White board

* Reflecting all colors.

* Good reflector, poor absorber.

White board: reflecting a lot of colors, absorbing just a little. The board is absorbing a little and the temperature is staying the same, so it must be emitting as much energy as it is absorbing. Since it is absorbing only a little bit → poor emitter.

Poor absorber = poor emitter

Doesn't absorb much = good reflector

Good absorber = good emitter

Doesn't reflect much = poor reflector

Blue marker: white light shining on it (incident on it) - molecules making up the paint are such that they absorb all the colors except for blue, which bounces off and we see that blue color.

Phone: detector of radiation.

* If we take this detector and detect the radiation someone emits, and the detector is connected to a computer and monitor, we can make a graph that detects energy emitted.

= Picking up the radiation that skin emits → graph shows energy emitted.

- Region of peak on graph = more energy is emitted by the skin (or some other surface of an object).

- Skin is at one temperature. One temperature at the surface gives rise to the object emitting energy of various wavelengths.

- Temperature: on the surface of the object.

If we look at the peak of the curve, where more energy is emitted, and find the wavelength associated with λₘₐₓ, that value is not the max wavelength emitted - it is the wavelength where more energy is emitted than any other wavelength → 1:1 correspondence.

☆☆☆ T in K ☆☆☆

YOU:

* Skin surface: T = 27 °C → 300 K

* Wavelength of energy = radiation emitted by our skin. Mostly, the energy is at what temperature?

* (λₘₐₓ) x (Tsᴜʀғᴀᴄᴇ)

Electromagnetic spectrum:

* Humans can only see wavelengths that fall in a certain range - can only detect from 700-400.

→ Depend on the light shining on us (from the cieling of the lecture room) to reflect back to the professor so he can see us).

→ We see most objects at room temperature, not by the radiation they emit. We see objects due to the visible light the objects reflect.

The calculation tells you the radiation emitted from the skin at room temperature, at this height.

To measure the temperature of the Sun, scientists aim a detector at the Sun → radiation hits the detector → detector is connected to a computer → graph is made → graph makes a curve → scientists find the peak and λₘₐₓ, which will be 500 nm from the Sun.

Between 400-700, the sun has a surface temperature at a certain value. This value allows the Sun to emit a lot of energy, precisely at wavelengths us humans can see. We can therefore figure out how hut the Sun is, because we know λₘₐₓ.

Tsᴜɴ = 5800 K = 5500 °C

* This process is how a thermometer measures our temperature to detect a fever.

During the day, us humans don't need to turn a light on because the Sun is hot enough that it emits enough energy of visible light. As our temperature goes up in value, the wavelength with which we begin to emit energy gets smaller and smaller. If we heat up to 5000 °C, we would be emitting energy.

The hotter a block of iron gets, the lower the wavelength with which it begins to emit. At room temperature, the iron emits various colors, but red begins to stand out the most when it is heated -→ then blue → etc.

EX: Northern lights.

Chapter 12: Thermodynamics

Mechanisms of Heat Transfer - Conduction, Convention and Radiation

If an object is at an absolute temperature T and its surroundings are at an absolute temperature Tₛ , then the net rate of energy transfer by the object as a result of thermal radiation is:

(dQ/dt)ₙₑₜ = 𝓸𝑒A(T⁴ - Tₛ⁴)

When an object is in thermal equilibrium with its environment, it radiates and absorbs energy at the same rate, and so (dQ/dt)ₙₑₜ = 0 and its temperature remains constant!

Chapter 12: Thermodynamics

Mechanisms of Heat Transfer - Conduction, Convention and Radiation

Thermos Bottles

A double-walled glass container with a vacuum between the walls. The glass walls are silvered to reflect radiation (sometimes these are called heat waves), and the vacuum prevents heat energy loss by conduction through the walls. If you keep the thermos bottle capped, then you prevent heat energy loss by convection.

Chapter 12: Thermodynamics

Mechanisms of Heat Transfer - Conduction, Convention and Radiation

Summary

Chapter 12: Thermodynamics

The Ideal Gas

Ideal gas model: a non-interacting gas and low density. There are two ways to write the equation of state of an ideal gas:

1. PV = nRT

P = pressure in the gas

V = volume occupied by the gas

n = number of moles of the gas

R = universal gas constant = 8.314 J/molK = 0.0821 atmL/molK

T = absolute temperature of the gas in Kelvin

One mole is that quantity of matter with Avogadro's number (Nᴀ = 6.02 x 10²³) of particles.

2. PV = NkʙT

N = total number of gas molecules (or atoms)

kʙ = Boltzmann constant = 1.38 x 10⁻²³ J/K

PIC: an idealized model of a gas

* Gas molecules are infinitely small.

* They exert forces on the walls of the container but not on each other.

Let...

m = mass of one molecule, and Mₘₒₗₑ = molar mass = mass of one mole of gas.

Then:

n = N/Nᴀ = Mₜₒₜₐₗ/Mₘₒₗₑ

also...

Mₘₒₗₑ = mNᴀ

R = Nᴀkʙ

If we want to raise the temperature of a gas, what (Q) do we use?

PV = nRT

☆☆☆ T in K ☆☆☆

State of gas: the particular situation the gas is in.

* EX: air in the lecture room.

- In a particular state, parameters (pressure, volume, temperature) are used to describe the state of the air.

- If volume, temperature, and moles of the air in the room are known, we can calculate the pressure of the air in the room.

- How can we change the state of the air in the room? We can shut the doors and start a fire - the temperature of the air will go up → not in its initial state anymore (new situation).

- In a particular state, parameters are also used to describe individual atoms: reach state where atom gets to lowest energy - ground state of Hydrogen atom: condition where the energy of an element is the lowest. If we shine a light on the Hydrogen atom → energy → promoted to new situation if the atom absorbs energy → in new state. In a new state, since the atom always wants to go back down to the ground state, it will go down to the ground state. As the atom changes from excited → ground, it is in different situations.

The transition from one state to another state is a process.

2nd expression for equation of state (ideal gas law): PV = NkʙT

* N represents something different - so its constant is different.

* EX: How many atoms of Hydrogen are there?

* Moles → # atoms there are.

1. Calculate the # of moles of gas → 2. Calculate the # of atoms of the gas.

The PV equation is easier to use to figure out the total # of gas molecules, rather than doing that Periodic Table calculation.

Ideal gas law:

* R = 8.314

Units: P = Pa, V = cm³, T = K, n = mol

- (Pa) x (m³) = J

* R = 0.0821

Units: P = atm, V = liters, T = K, n = mol

EX: Let's say you start a fire in lecture hall.

1. Fire → Temperature goes up → moving faster

2. Moving faster → hitting walls harder

3. Change in the temperature of the air → changing the state of the air (new temperature and pressure values).

The state of the system is represented by a point in a PV diagram.

* As air goes up in temperature, pressure goes up, so the state of the air changes → DIAGRAM IN PIC

- State: applies to a point on the graph.

- Process: applies to a line on the graph. The line describes a process.

Chapter 12: Thermodynamics

The Ideal Gas

xx

The state of a gas is represented in a Pressure versus Volume plot (or PV diagram) as a point on the graph.

A gas undergoes a change of state (or process) whenever the thermodynamic coordinates (like pressure, volume, temperature, etc) change in any way whatsoever. A process is thus described in a PV diagram as a curve. Let's consider the following processes that an ideal gas may be taken through:

(a) A process at Constant volume: (an isochoric or isovolumetric process)

Pₒ/Tₒ = P𝒻'/T𝒻

(b) A process at Constant pressure: (an isobaric process)

Vₒ/Tₒ = V𝒻'/T𝒻

(c) A process at Constant temperature: (an isothermal process)

PₒVₒ = P𝒻V𝒻

Diagrams: The equation of state applies to a point.

(a) Volume is constant, temperature ↑ = pressure ↑.

* Vᵢ = V𝒻

* Pᵢ ≠ P𝒻, so Tᵢ ≠ T𝒻

* If P and T at their initial states are known, and P at its final state is known, we can find T at its final state by:

→ Vᵢ = nRTᵢ/Pᵢ

→ Vᵢ = V𝒻

→ V𝒻 = nRT𝒻/P𝒻

→ T𝒻 = P𝒻V𝒻/nR

Vᵢ = equation of state, applies to initial state.

V𝒻 = equation of state, applies to final state.

(Tᵢ)/(Pᵢ) = (T𝒻)/(P𝒻) ← applies to PROCESS, not state.

☆☆☆ T by itself → K ☆☆☆

(b)

T𝒻 ≠ Tᵢ

* (Tᵢ)/(Vᵢ) = (T𝒻)/(V𝒻) ← does not apply to state - applies to process.

* Pᵢ & Tᵢ & Vᵢ : apply to a point, not a process.

(c) Tᵢ = constant.

Isochoric process → isochor

Isobaric process → isobar

Isothermal process → isotherm

(d) Adiatic: process where heat energy transfer = 0 (Q = 0).

You don't add heat or remove heat energy from gas. No heat energy enters or leaves gas.

Which curve is isothermal vs. adiabatic? Adiabats are steeper than isotherms.

MCAT: At what temperature does water have the highest density?

Chapter 12: Thermodynamics

Thermodynamics of an Ideal Gas

A. Work in Thermodynamics Processes

A system undergoes a change of state whenever the thermodynamic coordinates (like pressure, volume, temperature, etc) change in any way whatsoever.

1. Chemical equilibrium: when a system does not undego a change in internal structure such as a transfer of matter from one place to another.

2. Mechanical equilibrium: when a system does not experience a net (unbalanced) force or torque.

3. Thermal equilibrium: when all parts of a system are at the same temperature, and its temperature is the same as that of the surroundings.

When a system is in chemical, mechanical, and thermal equilibrium, then it is said to be in thermodynamic equilibrium. During a quasistatic process, the system is always infinitely near a state of thermodynamic equilibrium.

∑F = ma

Ideal gas:

* Disregard internal attractions between atoms and molecules making up the gas (they don't talk to one another, but they'll collide with one another as they move about).

* Disregard the volume occupied by each molecule - as if the entire volume is accessible to each molecule making up the gas.

* Calculate internal energy (Eₗₙₜ) as if molecules in the air do not interact with each other.

Van der Waals Forces: take into account the interactions between gas molecules and the fact that they do take up space.

Monatomic gas: disregarding interactions between molecules:

kinetic energy + kinetic energy + ... (of each atom) → Eₗₙₜ

∑KE = Eₗₙₜ

Montatomic gas: not the same mass (m),

Eₗₙₜ = (1/2)(m₁v₁)² + (1/2)(m₂v₂)² + ...

V(ɢᴀs): if calculating the average KE of each molecule:

* M = mass of each particle

* V = volume

* T = temperature

* P = pressure

* T: a measure of the average KE of each constituent.

* Ideal gas (monatomic)

→ What must change if the Eₗₙₜ changes? (assume no adding/removal of molecules) → Temperature!

If temperature doesn't change, the ΔEₗₙₜ = 0.

If ΔEₗₙₜ = 0, the temperature is constant (isothermal process).

☆☆☆ Know ΔEₗₙₜ = 0 for isothermal process! ☆☆☆

If ΔEₗₙₜ = 0, then the work done on the gas is zero.

→ If ΔEₗₙₜ = 0, then W(ᴅᴏɴᴇ ᴏɴ ɢᴀs) = 0.

Chapter 12: Thermodynamics

Thermodynamics of an Ideal Gas

A. Work in Thermodynamics Process

v

The work W done on a gas as its volume changes from some initial value Vᵢ, to some final value V𝒻 is given by:

Wₒₙ ₜₕₑ 𝓰ₐₛ = PIC

where P = the pressure in the gas which may vary during the process.

Compression work done on the gas (W > 0).

Expansion work done by the system (W < 0).

The work done on the gas is equal to the:

a) (- area ) under the PV curve if the gas expands, or

b) ( + area ) under the PV curve if the gas contracts.

Chapter 12: Thermodynamics

Thermodynamics of an Ideal Gas

A. Work in Thermodynamics Process

vv

Let's consider the work done on a gas during different quasistatic processes:

(a) Constant volume: (an isochoric or isovolumetric process)

- dV = 0 therefore W = 0

(area under the curve is zero)

(b) Constant pressure: (an isobaric process)

W = -(area under the curve)

= - P (V𝒻' - Vₒ)

(c) Constant temperature: an isothermal process.

For an ideal gas, the equation of state is PV = nRT,

W = PIC

Chapter 12: Thermodynamics

Thermodynamics of an Ideal Gas

B. Internal Energy Eₗₙₜ

Internal Energy: the energy associated with the atoms and molecules of a system. It includes the kinetic and potential energies associated with the random translational, rotational, and vibrational motions of the atoms or molecules that make up the system, as well as the intermolecular potential energies.

The total translational kinetic energy of a system of molecules is proportional to the absolute temperature of the system,

Eₗₙₜ = N[(1/2)(mv²)] = (3/2)(NkʙT) = (3/2)nRT

For a monatomic gas, the translational kinetic energy is the only type of energy the molecule can have. Thus, the above expression equals the internal energy Eₗₙₜ of a monatomic gas.

Eₗₙₜ = (3/2)(nRT) (ideal monatomic gas)

More generally, the internal energy of an ideal gas may be written as...

Eₗₙₜ = nCᵥT (ideal gas)

Note that the internal energy of an ideal gas depends only on its temperature, not on its pressure or volume.

Chapter 12: Thermodynamics

Thermodynamics of an Ideal Gas

C. Heat Capacities of Gases

Cᵥ = molar specific heat at constant volume

Cₚ = molar specific heat at constant pressure

Modify the equation Q = mcΔT which was good for solids and liquids, and write this equation for gases as:

Q = nCᵥΔT (for an isochoric process)

Q = nCₚΔT (for an isobaric process)

Cᵥ and Cₚ are related by: Cₚ = Cᵥ + R γ = (Cₚ/Cᵥ)

(a) for monatomic gases (He, Ar, Ne):

Cᵥ = (3/2)R

Cₚ = (5/2)R

(b) for diatomic gases (H₂, N₂, O₂, CO):

Cᵥ = (5/2)R

Cₚ = (7/2)R

PIC: Experimental values of Cᵥ the molar heat capacity at constant volume for hydrogen gas (H₂). The temperature is plotted on a logarithmic scale.

* Below 50 K, H₂ molecules undergo translation but do not rotate or vibrate.

* Appreciable rotational motion begins to occur above 50 K.

* Appreciable vibrational motion begins to occur above 600 K.

Chapter 12: Thermodynamics

Thermodynamics of an Ideal Gas

D. The First law of Thermodynamics

ΔEₗₙₜ = W + Q

W = work done on the system

Q = heat energy entering the system is positive

ΔEₗₙₜ = change in internal energy of the system