Thermodynamics and Heat Transfer

Questions on Thermal Equilibrium and Temperature Measurement

  • Question 1: What does it mean to say that two systems are in thermal equilibrium?
    • Definition of Thermal Equilibrium: Two systems are in thermal equilibrium when there is no net flow of thermal energy between them, meaning both systems have reached the same temperature. This concept is governed by the Zeroth Law of Thermodynamics, which states that if system A is in thermal equilibrium with system B, and system B is in thermal equilibrium with system C, then system A is also in thermal equilibrium with system C.

Physical Properties and Temperature Measurement

  • Question 2: Give an example of a physical property that varies with temperature and describe how it is used to measure temperature.
    • Example of a Physical Property: One prominent example of a property that varies with temperature is the volume of a liquid.
    • Measurement of Temperature Using Volume: A common application of this principle is found in alcohol thermometers, where the volume of alcohol expands or contracts in response to temperature changes. The level of alcohol in the calibrated tube provides a direct reading of the temperature.

Thermometers and Temperature Dynamics

  • Question 3: When a cold alcohol thermometer is placed in a hot liquid, the column of alcohol goes down slightly before going up. Explain why.
    • Explanation of Alcohol Level Dynamics: Initially, when the cold alcohol thermometer is placed into a hot liquid, the alcohol in the thermometer is at a lower temperature than the hot liquid. As the alcohol warms up, it contracts briefly due to thermal inertia and thermal lag. After this initial contraction, as the alcohol absorbs heat energy from the hot liquid, it eventually expands, causing the column of alcohol to rise.

Equilibrium Temperature Considerations

  • Question 4: If you add boiling water to a cup at room temperature, what would you expect the final equilibrium temperature of the unit to be? Consider the surroundings as part of the system.
    • Answer: The final equilibrium temperature will stabilize somewhere between the boiling point of water (100ºC) and the room temperature (approximately 20-25ºC). This outcome considers the heat exchange with the environment; heat will flow from the boiling water to the cooler components (cup and surroundings) until thermal equilibrium is reached. The specific final value depends on heat capacities of the water, cup, and surrounding environment, along with their volumes.

Thermal Stresses and Material Properties

  • Question 5: Explain why Pyrex®, a glass with a small coefficient of linear expansion, is less susceptible to thermal stresses caused by uneven cooling which can break glass cookware.
    • Explanation of Pyrex's Resistance to Thermal Stresses: Pyrex® glass is engineered to have a low coefficient of linear expansion, meaning it expands and contracts less in response to temperature changes than regular glass. Therefore, when subjected to sudden temperature shifts, Pyrex® can withstand these stresses better than traditional glass, significantly reducing the risk of cracking or breaking.

Implications of Water's Anomalous Expansion

  • Question 6: Discuss the implications of cell damage due to the expansion of water when freezing for the prospect of preserving human bodies by freezing.
    • Discussion on Cell Damage: Water freezes with a significant volume increase (about 9%), which can result in the formation and growth of ice crystals that rupture cellular structures. In biological preservation, this damage prompts critical questions about the feasibility of cryopreservation techniques for humans or other organisms. If a high percentage (10%-30%) of cells can potentially burst during freezing, it raises ethical and functional considerations on the reliability of reviving a preserved organism without severe cellular damage.

Thermal Expansion and Insertion Techniques

  • Question 7: Should the block be hotter or colder than the peg during insertion to obtain a tight fit?
    • Explanation: The block should be hotter than the peg during insertion. As the block (which must be heated) expands more when warmed, it allows the slightly larger peg to fit into it without resistance. Upon cooling, the block will contract, securing the peg tightly in place.

Practical Applications of Heat Transfer

  • Question 8: Does running hot water over a tight metal lid on a glass jar help to open it? Explain.
    • Explanation of Heat Transfer Efficacy: Yes, running hot water over a tight metal lid helps by the principle of thermal expansion. The metal lid expands when heated, which can create enough of a gap to break the seal with the jar, enabling easier removal. Conversely, glass does not expand as much and remains relatively unchanged, facilitating the separation between lid and jar.

Population Calculations

  • Question 9: Find out the human population of Earth. Is there a mole of people inhabiting Earth? If the average mass of a person is 60 kg, calculate the mass of a mole of people. How does the mass of a mole of people compare with the mass of Earth?
    • Current Human Population Estimate: Approximately 7.9 billion people.
    • Calculating Mass of a Mole of People: A mole contains approximately $6.022 imes 10^{23}$ entities. To find the mass of a mole of people:
    • ext{Mass of a mole of people} = 60 ext{ kg/person} imes 6.022 imes 10^{23} ext{ people} = 3.613 imes 10^{25} ext{ kg}
    • Comparison with Earth: The mass of the Earth is about $5.972 imes 10^{24} ext{ kg}$. Therefore, the mass of a mole of people is approximately 6.05 times greater than the total mass of the Earth, indicating that the number of people on Earth is less than a mole.

Temperature Conversions

  • Question 10: What is the Fahrenheit temperature of a person with a 39.0ºC fever?
    • Fahrenheit Conversion Formula: The conversion from Celsius to Fahrenheit is given by:
    • F = (C imes rac{9}{5}) + 32
    • For 39.0ºC:
      • F = (39.0 imes rac{9}{5}) + 32 = 102.2ºF

Thermometer Shielding

  • Question 22: Why are thermometers that are used in weather stations shielded from sunshine? What does a thermometer measure if it is shielded from the sunshine and also if it is not?
    • Reason for Shielding: Thermometers are shielded from direct sunlight to prevent radiative heating, which would lead to inaccurate temperature readings.
    • Measured Temperature with Shielding: When shielded, the thermometer measures the air temperature more accurately, reflecting environmental conditions. Without shielding, it may measure elevated temperatures due to solar radiation, affecting reliability in climate data collection.

Net Heat Transfer Calculation

  • Question 23: On a hot day, the temperature of an 80,000-L swimming pool increases by 1.50ºC. What is the net heat transfer during this heating? Ignore complications like evaporative loss.
    • Calculation of Heat Transfer: The heat transfer can be calculated using the formula:
    • Q = mc riangle T
    • Where:
    • $m = 80,000 ext{ kg}$ (assuming the density of water is about 1 kg/L),
    • $c = 4.186 ext{ kJ/(kg C)}$,
    • $ riangle T = 1.50 ext{ C}$.
    • Thus,
    • Q = 80,000 ext{ kg} imes 4.186 ext{ kJ/(kg C)} imes 1.50 ext{ C} = 502.32 ext{ kJ}

Temperature Conversions and Scale References

  • Question 11: Frost damage to most plants occurs at temperatures of 28.0ºF or lower. What is this temperature on the Kelvin scale?
    • Temperature Conversion: To convert Fahrenheit to Kelvin, use the formula:
    • K = (F - 32) imes rac{5}{9} + 273.15
    • For 28.0ºF:
    • K = (28.0 - 32) imes rac{5}{9} + 273.15 = 260.37 ext{ K}

Energy Conservation and Temperature Settings

  • Question 12: To conserve energy, room temperatures are kept at 68.0ºF in winter and 78.0ºF in summer. What are these temperatures on the Celsius scale?
    • Celsius Conversion Formula:
    • C = (F - 32) imes rac{5}{9}
    • Room temperature in winter:
    • C_{winter} = (68.0 - 32) imes rac{5}{9} = 20.0ºC
    • Room temperature in summer:
    • C_{summer} = (78.0 - 32) imes rac{5}{9} = 25.56ºC

Temperature Conversions for Electrical Equipment

  • Question 13: A tungsten light bulb filament may operate at 2900 K. What is its Fahrenheit and Celsius temperature?
    • Celsius Conversion:
    • C = K - 273.15 = 2900 - 273.15 = 2626.85ºC
    • Fahrenheit Conversion:
    • F = (C imes rac{9}{5}) + 32 = (2626.85 imes rac{9}{5}) + 32 = 4779.33ºF

Thermal Expansion Impact on Structures

  • Question 14: The height of the Washington Monument is measured to be 170 m on a day when the temperature is 35.0ºC. What will its height be on a day when the temperature falls to -10.0ºC?
    • Height Change Calculation: Using the formula:
    • riangle L = L_0 imes eta imes riangle T
    • Assuming a thermal expansion coefficient ($eta$) of limestone is approximately $7.5 imes 10^{-6} ext{ °C}^{-1}$, and the original height $L_0 = 170 ext{ m}$:
    • Change in temperature:
    • $ riangle T = (-10.0 - 35.0) = -45.0ºC$
    • Thus, the height change becomes:
    • riangle L = 170 imes (7.5 imes 10^{-6}) imes (-45.0) = -0.05737 ext{ m}
    • The new height is:
    • 170 ext{ m} - 0.05737 ext{ m} = 169.94263 ext{ m}

Temperature-Related Expansion in Structures

  • Question 15: How much taller does the Eiffel Tower become if the temperature increases by 15ºC? The original height is 321 m and it's made of steel.
    • Calculation of Height Change: Assuming a thermal expansion coefficient for steel is approximately $11 imes 10^{-6} ext{ °C}^{-1}$, we calculate:
    • Height change:
    • riangle L = L_0 imes eta imes riangle T
    • So,
    • riangle L = 321 ext{ m} imes (11 imes 10^{-6}) imes (15) = 0.053 ext{ m}
    • Therefore, the height of the Eiffel Tower increases to:
    • 321 ext{ m} + 0.053 ext{ m} = 321.053 ext{ m}

Instant Coffee Cooling Analysis

  • Question 16: As a physicist makes a cup of instant coffee and notices that, as the coffee cools, its level drops 3.00 mm in the glass cup, showing this decrease cannot be due to thermal contraction.
    • Volume Calculation for Decrease: Given:
    • Volume of coffee = 350 cm³.
    • Diameter of cup = 7.00 cm, so radius = 3.50 cm. The cross-sectional area (A) = rac{ ext{Area} = ext{Diameter}^2 imes rac{ ext{π}}{4}} = rac{(7.00)^2 imes rac{ ext{π}}{4}}
    • Now, the volume change due to cooling segment can be calculated:
    • V = A imes h = ext{Area} imes 3.00 ext{ mm/m} ext{ (converted to cm)}
    • Using the above values establishes that the volume change attributed to contraction is not congruent with the drop noted, leading to the conclusion that air escape is primarily the factor behind the drop in coffee level.

Pressure and Temperature in Tires

  • Question 17: The gauge pressure in car tires is 2.50 × 10^5 N/m² at a temperature of 35.0° C, what is the pressure at -40.0° C?
    • Calculation using Ideal Gas Law: Applying Gay-Lussac's Law, where pressure is proportional to temperature (at constant volume), we calculate:
    • rac{P1}{T1} = rac{P2}{T2}
    • Where:
    • $P_1 = 2.50 × 10^5 N/m²$,
    • $T_1 = 35.0 + 273.15 = 308.15 K$,
    • $T_2 = -40.0 + 273.15 = 233.15 K$.
    • Rearranging gives us:
    • P2 = P1 imes rac{T2}{T1}
    • P_2 = 2.50 imes 10^5 N/m^2 imes rac{233.15}{308.15} = 1.89 imes 10^5 N/m^2

Heat Transfer Correlations

  • Question 18: How is heat transfer related to temperature?
    • Correlation: Heat transfer is fundamentally linked to the temperature difference between objects or systems; heat flows naturally from hot regions to cold regions until thermal equilibrium is achieved. It follows the Second Law of Thermodynamics, emphasizing that due to this drive toward equilibrium, the entropy of a system tends to increase overall.

Situations of Heat Transfer

  • Question 19: Describe a situation in which heat transfer occurs and the resulting forms of energy.
    • Example Situation: A hot cup of coffee sitting on a table. Heat is transferred from the coffee (hot region) to the surrounding air (cooler environment), which leads to the coffee cooling down and air heating up slightly.
    • Resulting Forms of Energy: The energy forms involved here are thermal energy exchanged between coffee and air, leading to decreased internal energy within the coffee and slight increases in the internal energy of the surrounding air.

Braking and Energy Transfer

  • Question 20: The brakes in a car increase in temperature by ΔT when bringing the car to a stop from speed v. How much greater would ΔT be if the car initially had twice the speed?
    • Explanation: The energy dissipated by brakes during stopping relates to the kinetic energy equation:
    • KE = rac{1}{2}mv^2
    • If the speed doubles ($v = 2v$), then:
    • The new kinetic energy becomes $KE' = rac{1}{2}m(2v)^2 = 2^2 imes rac{1}{2}mv^2 = 4 imes KE$.
    • Thus, the energy required to stop the vehicle would result in a temperature increase ($ riangle T$) that is four times greater than initially required.

Cloudy vs Clear Nights

  • Question 21: Why are cloudy nights generally warmer than clear ones?
    • Explanation: Cloud cover acts as an insulating layer that traps heat radiating from the Earth's surface. During a clear night without clouds, heat escapes easily into space, leading to colder temperatures. In contrast, clouds help retain some of that heat, thus resulting in comparatively warmer nighttime temperatures.

Heat Transfer for Sterilization

  • Question 24: To sterilize a 50.0-g glass baby bottle, we must raise its temperature from 22.0ºC to 95.0ºC. How much heat transfer is required?
    • Calculation of Heat Transfer:
    • Q = mc riangle T
    • For a glass bottle ( ext{specific heat} ext{ of glass } ext{≈} 0.84 ext{ J/(g} ext{ºC)}):
    • Mass (m) = 50.0 g,
    • Change in temperature ($ riangle T$) = 95.0 - 22.0 = 73.0ºC
    • Therefore:
    • Q = 50.0 ext{ g} imes 0.84 ext{ J/(g} ext{ºC)} imes 73.0ºC = 3066 ext{ J}

Work and Energy Conversion

  • Question 25: Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 7.50 cm per rub, and with an average frictional force of 40.0 N, what is the temperature increase?
    • Total Work Done Calculation:
    • Work ($W$) = force × distance = $F imes d$. For 20 rubs,
    • W = (40.0 ext{ N}) imes (20 imes 7.50 ext{ cm} imes 0.01 ext{ m/cm})
    • W = (40.0) imes (1.50) = 60 ext{ J}
    • Temperature Change: Given the specific heat of tissue ($c ext{ ≈ } 3.5 ext{ J/(g} ext{ºC)}$ and mass warmed = 0.100 kg = 100g):
    • riangle T = rac{W}{mc} = rac{60 ext{ J}}{100 ext{ g} imes 3.5 ext{ J/(g} ext{ºC)} } ext{ = } 0.171ºC

Specific Heat Exploration

  • Question 26: A block of pure material (0.250 kg) is heated from 20.0ºC to 65.0ºC by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance.
    • Specific Heat Calculation:
    • Using $Q = mc riangle T$, where
    • $Q = 4350 ext{ J}$,
    • m = 0.250 kg,
    • $ riangle T = 65.0 - 20.0 = 45.0ºC$:
    • Therefore, we can rearrange to find c:
    • c = rac{Q}{m imes riangle T} = rac{4350 ext{ J}}{0.250 ext{ kg} imes 45.0 ext{ ºC}}
    • c = 0.386 ext{ J/(g} ext{ºC)}
    • Upon comparison, this specific heat corresponds closely to Aluminum, suggesting the substance is likely made of pure aluminum.

Copper vs Water Heat Transfer Ratios

  • Question 27: Suppose identical amounts of heat transfer into different masses of copper and water cause identical changes in temperature. What is the ratio of the mass of copper to water?
    • Heat Transfer Analysis: Since the heat transfer is the same, we apply $Q = mc riangle T$:
    • Let’s denote mass of water = mw, mass of copper = mc. Since $c{water} ext{ ≈ } 4.186 ext{ J/(g} ext{ºC)}$ and $c{copper} ext{ ≈ } 0.385 ext{ J/(g} ext{ºC)}$, then:
    • Given identical $ riangle T$:
    • mw imes 4.186 imes riangle T = mc imes 0.385 imes riangle T
    • Solving yields:
    • rac{mc}{mw} = rac{4.186}{0.385} ext{ thus } mc = 10.86 mw
    • The mass of copper will be 10.86 times greater than the mass of water.

Kinetic and Gravitational Energy Interactions

  • Question 28: If you run down some stairs and stop, what happens to your kinetic energy and initial gravitational potential energy?
    • Energy Transformation: As you run down the stairs, your gravitational potential energy is converted into kinetic energy; as you come to a stop, your kinetic energy is dissipated, primarily as thermal energy due to friction and air resistance, thus fulfilling the conservation of energy principle—energy is neither created nor destroyed but changes forms.

Kinetic Energy Transfer Types

  • Question 29: Identify the type of energy transferred to your body in various situations as either internal energy, heat transfer, or doing work.
    • (a) Basking in sunlight: Heat Transfer (heat from the sun warms the body)
    • (b) Eating food: Internal Energy (chemical energy from food is converted into internal energy)
    • (c) Riding an elevator to a higher floor: Doing Work (mechanical work is performed on the body to elevate it)

Change in Internal Energy for Gasoline

  • Question 30: What is the change in internal energy of a car if you put 12.0 gal of gasoline into its tank?
    • Calculation of Internal Energy Change: Given the energy content of gasoline is approximately $1.3 imes 10^8 ext{ J/gal}$, changing energy when 12.0 gal is added becomes:
    • ext{Change in internal energy} = 12.0 imes (1.3 imes 10^8) = 1.56 imes 10^9 ext{ J}, implying significant energy increase solely based on gasoline volume.

Heat Transfer Calculations

  • Question 31: How much heat transfer occurs from a system if its internal energy decreased by 150 J and while doing 30.0 J of work?
    • Energy Calculation: The First Law of Thermodynamics indicates that:
    • riangle U = Q - W
    • Rearranged for Q gives:
    • Q = riangle U + W
    • Plugging values:
    • Q = -150 ext{ J} + 30.0 ext{ J} = -120 ext{ J}
    • Therefore, 120 J of heat is released from the system.

Change in Internal Energy Exiting the System

  • Question 32: A system does $1.80 imes 10^8 ext{ J}$ of work while $7.50 imes 10^8 ext{ J}$ of heat transfer occurs to the environment. What is the change in internal energy of the system?
    • Using the First Law:
    • riangle U = Q - W
    • Here,
    • riangle U = -7.50 imes 10^8 ext{ J} - 1.80 imes 10^8 ext{ J}
    • Therefore,
    • riangle U = -9.30 imes 10^8 ext{ J} indicating significant energy loss from the system.

Frequency Changes in Material Types

  • Question 33: Explain why you expect an object made of a stiff material to vibrate at a higher frequency than a similar object made of a spongy material.
    • Explanation: Stiffer materials possess a greater restoring force when subjected to disturbance, resulting in quicker returns to equilibrium position post-disturbance. Consequently, this increases the vibrational frequency compared to spongy materials, which experience significant deformation and slow recovery during oscillations.

Trailer Behavior on the Road

  • Question 34: As you pass a freight truck with a trailer on a highway, you notice that its trailer is bouncing up and down slowly. Is it more likely that the trailer is heavily loaded or nearly empty? Explain.
    • Explanation of Load Condition: A trailer that bounces slowly suggests it is nearly empty; an overload would lead to faster oscillations, given increased mass would dampen the upward/downward motion rates compared to a light and less damped trailer.

Performance Enhancements for Modified Vehicles

  • Question 35: Some people modify cars to be much closer to the ground than originally manufactured. Should they install stiffer springs?
    • Answer: Yes, they should install stiffer springs. A lower ride height results in a loss of suspension travel, requiring stiffer springs to maintain proper vehicle handling, stability, and to avoid excessive bottoming out during bumps or irregularities on the road.

Pendulum Length in Different Gravitational Fields

  • Question 36: If you move to a city with a slightly greater acceleration due to gravity and take your pendulum clock with you, will you have to lengthen or shorten the pendulum to keep the correct time, other factors remaining constant?
    • Explanation: You will need to shorten the pendulum. A pendulum's period increases (i.e., it swings slower) with lengthening, and with a stronger gravitational pull, the swing speed increases requiring a shorter effective length to maintain the existing timing mechanism.

Wave Type Examples

  • Question 37: Give one example of a transverse wave and another of a longitudinal wave, being careful to note the relative directions of the disturbance and wave propagation in each.
    • Transverse Wave Example: Light waves are transverse waves where the disturbance is perpendicular to the direction of wave propagation.
    • Longitudinal Wave Example: Sound waves in air illustrate longitudinal waves, where the disturbance happens in the same direction as wave propagation (e.g., air compressions and rarefactions).

Speaker Connection Impact

  • Question 38: Explain why it is important to have both speakers connected the same way.
    • Explanation: Connecting speaker wires incorrectly can cause them to move out of phase, leading to destructive interference, canceling sounds and compromising audio clarity. Proper connections ensure that both speakers push and pull in synchrony, delivering accurate sound reproduction.

Spring Scale Calculations

  • Question 39: Fish are hung on a spring scale to determine their mass.

    • (a) What is the force constant of the spring if it stretches 8.00 cm for a 10.0 kg load?

    • F = kx where $F$ is the weight of the load $= mg$, thus the force constant k = rac{F}{x} = rac{(10.0 ext{ kg } imes 9.81)}{0.08 ext{ m}} = 1226.25 ext{ N/m}

    • (b) What is the mass of a fish that stretches the spring 5.50 cm?

    • ext{Force due to fish mass} = kx

    • Rearranging gives:

    • m = rac{kx}{g} = rac{1226.25 ext{ N/m } imes 0.055 ext{ m}}{9.81} ext{ = 7.05 kg}

    • (c) How far apart are the half-kilogram marks on the scale?

    • For every half-kilogram increase:

    • 4.905 ext{ N} = k ext{(half kg)} o x = rac{4.905}{1226.25} ext{ m} = 0.00402 ext{ m} = 4.02 cm

Bathroom Scale Mechanics

  • Question 40: If the bathroom scale used to assess eligibility can be described by Hooke’s law and is depressed 0.75 cm by its maximum load of 120 kg:

    • (a) What is the effective spring constant?

    • k = rac{F}{x} = rac{(120 ext{ kg } imes 9.81)}{0.0075 ext{ m}} = 1.047 imes 10^5 ext{ N/m}

    • (b) A player stands on the scales and depresses it by 0.48 cm. Is he eligible to play on this under-85 kg team?

    • Weight: F = kx = 1.047 imes 10^5 ext{ N/m} imes 0.0048 ext{ m} = 502.56 ext{ N}

    • ext{mass} = rac{502.56}{9.81} = 51.1 ext{ kg} indicating eligible.

Truck Spring Constant Calculation

  • Question 41: The springs of a pickup truck act like a single spring with a force constant of $1.30 imes 10^5$ N/m. By how much will the truck be depressed by its maximum load of 1000 kg?
    • Depression Calculation:
    • F = mg = 1000 ext{ kg} imes 9.81 ext{ m/s}^2 = 9810 ext{ N}
    • Using Hooke’s law:
    • x = rac{F}{k} = rac{9810 ext{ N}}{1.30 imes 10^5 ext{ N/m}} = 0.0756 ext{ m} = 7.56 ext{ cm}

Tuning Fork Frequency Calculation

  • Question 42: Find the frequency of a tuning fork that takes $2.50 imes 10^{-3}$ s to complete one oscillation.
    • Frequency ($f$) Calculation:
    • f = rac{1}{T} = rac{1}{2.50 imes 10^{-3}} = 400 ext{ Hz}

Cuckoo Clock Spring Constants

  • Question 43: What force constant is needed to produce a period of 0.500 s for a 0.0150 kg mass in a cuckoo clock?
    • Spring Constants via Period Formula:
    • The period of oscillation for a mass on a spring is given by:
    • T = 2 ext{π} imes ext{} rac{m}{k}
    • Rearranging for k gives:
    • k = rac{4 ext{π}^2m}{T^2} = rac{4 ext{π}^2 (0.0150)}{(0.500)^2} ext{ = } 0.3 ext{ N/m}

Change of Oscillation Period Calculation

  • Question 44: A 0.500 kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to change the period to 2.00 s?
    • Calculating New Mass Requirement: The period of a mass-spring system depends on both mass and spring constant. We know:
    • T = 2 ext{π} imes rac{m}{k}
    • By maintaining the spring constant constant and keeping $T$ as 2.00 s, you need to solve:
    • rac{T1^2}{m1} = rac{T2^2}{m2} leads to:
    • With mass $m1 = 0.500 ext{ kg}, T1= 1.50 ext{ s}, T_2 = 2.00 ext{ s},$ we rearrange to find new mass:
    • Thus $m2 = m1 o ext{derived mass ratio leads to} 0.500 ext{ kg} o k$ where k is spring invariant leading new mass contribution.

Pendulum Length Calculation

  • Question 45: The length of a pendulum that has a period of 1.00 s can be associated with its dynamics according to the formula:
    • T = 2 ext{π} imes rac{L}{g} leading to L = rac{gT^2}{4 ext{π}^2}
    • With $T = 1.0 s$, let's estimate:
    • L = g T^2 rac{ ext{ in meters} = 9.81 imes (1.00)^2}{4 ext{π}^2}
    • Hence, resulting in about $L = 0.25 m$ as the pendulum length approximating nature's principle.

Pendulum Period Calculation

  • Question 46: What is the period of a 1.00 m long pendulum?
    • Calculation of Period: Using:
    • T = 2 ext{π} imes rac{L}{g} = 2 ext{π} imes rac{1.00 ext{ m}}{9.81 ext{ m/s}^2}
    • Thus, $T ext{ ≈ } 2.01 ext{ s}.$

Community Concerns Regarding Sound Levels

  • Question 47: A community is concerned about a plan to bring train service to their downtown. The current sound intensity level is 70 dB. If the mayor assures the public that there will be a difference of only 30 dB in sound in the area, should the townspeople be concerned? Why?
    • Concern Justification: The decibel scale is logarithmic. An increase of 30 dB means the sound intensity will increase significantly more than merely doubling (as it represents). The final sound intensity would therefore be 100 dB, a considerable increase that can cause annoyance or health issues, implying that the interpretation of sound impacts is significant—and townspeople ought to indeed be concerned.

Sound Wave Calculations

  • Question 48: When poked by a spear, an operatic soprano lets out a 1200-Hz shriek. What is its wavelength if the speed of sound is 345 m/s?
    • Applying Wave Equation:
    • v = f imes ext{wavelength} = 345 m/s = 1200 Hz imes ext{wavelength} leads to
    • Rearrangement gives $ ext{wavelength} = rac{345}{1200} = 0.2875 m$ or roughly 28.75 cm.

Sound Frequency Determination

  • Question 49: What frequency sound has a 0.10-m wavelength when the speed of sound is 340 m/s?
    • Frequency Calculated Using Wave Equation:
    • f = rac{v}{ ext{wavelength}} = rac{340}{0.10} = 3400 Hz

Speed of Sound Calculation

  • Question 50: Calculate the speed of sound on a day when a 1500 Hz frequency has a wavelength of 0.221 m.
    • Wave Speed Calculation:
    • Speed of sound v = f imes ext{wavelength} = 1500 Hz imes 0.221 m = 331.5 m/s

Sound Speed in High Temperature Environments

  • Question 51: In the Sahara Desert, air temperatures can reach 56.0ºC. What is the speed of sound at that temperature?
    • Speed of Sound Calculation Relation to Temperature: The speed of sound can be found with the approximation:
    • v ext{ (m/s)} = 331 + 0.6T where T is in Celsius.
    • So at 56.0ºC, then:
    • v = 331 + 0.6 imes 56.0 = 331 + 33.6 = 364.6 ext{ m/s}

Energy Production and Sound Intensity Levels

  • Question 52: The warning tag on a lawn mower states it produces noise at a level of 91.0 dB. What is this in watts per meter squared?
    • Sound Intensity Calculation Using dB Scale:
    • The relationship is given by:
    • L = 10 ext{ log}{10} rac{I}{I0} where $I_0 = 1.0 imes 10^{-12} W/m^2$ is the reference intensity.
    • Thus, converting the dB level to intensity:
    • Rearranging yields,
    • I = I_0 imes 10^{(L/10)} = 1.0 imes 10^{-12} imes 10^{(91/10)} = 1.26 imes 10^{-5} W/m^2

Intensity Levels of Earphones

  • Question 53: What sound intensity level in dB is produced by earphones that create an intensity of $4.00 imes 10^{-2} W/m^2$?
    • Intensity Level Calculation:
    • L = 10 ext{ log}{10} rac{I}{I0} = 10 ext{ log}_{10} rac{4.00 imes 10^{-2}}{1.0 imes 10^{-12}}
    • Evaluating the expression yields:
    • Thus, $L ext{ ≈ } 16.24 ext{ dB}$.

Comparing Intensity Levels

  • Question 54: (a) What is the decibel level of a sound that is twice as intense as a 90.0-dB sound? (b) What about one-fifth as intense?
    • (a) Doubling Intensity:
    • Increase of approx. 3 dB,
    • L_{ ext{new}} = 90.0 + 3 = 93.0 ext{ dB}
    • (b) Dividing Intensity: The corresponding drop leads to about 7 dB, thus:
    • L_{ ext{new}} = 90.0 - 7 = 83.0 ext{ dB}

Calculating Differences in Sound Intensity Levels

  • Question 55: (a) What is the intensity of a sound that has a level 7.00 dB lower than a $4.00 imes 10^{-9} W/m^2$ sound? (b) What is the intensity of a sound that is 3.00 dB higher?
    • (a) Intensity Level Lowering:
    • L = 10 ext{ log}{10} rac{I}{I0} = L_0 - 7 = 4.00 imes 10^{-9} ext{ W/m}^2 gives $I = 2.00 imes 10^{-10} W/m^2$
    • (b) Intensity Level Increase:
    • L = 4.00 imes 10^{-9} + 3 = 4.12 imes 10^{-9}W/m^2

More Intensity Level Calculations

  • Question 56: (a) How much more intense is a sound that has a level 17.0 dB higher than another? (b) If one sound has a level 23.0 dB less than another, what is the ratio of their intensities?
    • (a) Intensity Increase Calculation: Each 10 dB increases the sound intensity by a factor of 10, hence 17 dB reflects a 50 times greater intensity scale.
    • (b) Intensity Comparison: A 10 dB drop signifies lesser intensity by a factor of 10, therefore; a 20 dB difference implies 100 times lower intensity.

Additional Notes on Temperature Measures

  • Temperature Dynamics:

    • Everyday Context: It reflects sensations of hot and cold, derived from measuring the average kinetic energy at a molecular level.

  • Temperature Scales:

    • Celsius, Fahrenheit, Kelvin: Clearly differentiated contexts of temperature norms and measurement systems in use globally.

  • Temperature Conversions:

    • Fahrenheit to Celsius: C = rac{5}{9}(F - 32)
    • Kelvin to Celsius: K = C + 273.15

Thermal Conductivity and Expansion Concepts

  • Thermal Equilibrium: Defined where no heat flow occurs, indicating equal temperatures are achieved through thermal interactions.

  • Thermal Conductivity: Reflecting an object's capacity to conduct heat; impacted strongly by material properties.

  • Heat Transfer Mechanisms:

    • Conduction (direct contact), Convection (molecular movement), Radiation (energy carried through space).

  • Thermal Expansion:

    • Linear Expansion Formula: riangle L = etaL_0 riangle T
    • Volume Expansion Formula: riangle V = eta'V_0 riangle T
    • Where coefficients signify inverse temperature units. Positive values indicate expansion; negative, contraction.

  • Anomalous Expansion: Referring to uncommon expansion/contraction behavior of materials, especially water's unusual behavior around 4ºC to ice formation.

Gas Laws Overview

  • Gas Principles: Relate pressure and volume effects, emphasizing proportionality factors with temperature regulation context in thermodynamics.
    • Boyle’s Law: Volume inversely proportional to pressure at constant temperature.
    • Charles's Law: Volume proportional to temperature at constant pressure.
    • Gay-Lussac’s Law: Pressure is proportional to temperature at constant volume.

Mole Concept

  • Mole Definition: The amount of substance whose mass in grams equals its molecular mass, integrating Avogadro’s number of approximately $6.022 imes 10^{23}$ molecules.

Heat Units Definition

  • Heat Conversion Units:
    • One calorie (cal) = 4.184 J.
    • One kilocalorie = 4.184 kJ.
    • A dietary calorie (Cal) corresponds to one kilocalorie.

Heat Capacity and Related Thermodynamic Principles

  • Specific Heat Equation: Q = mc riangle T

    • Where specific heat c is the substance-dependent heat required to alter mass temperature. Defined in J/(kg•ºC).

  • Latent Heat for Phase Changes: Energy relationships under phase transitions measured through Q = mL where L is the latent heat associated with the transition dynamics.

  • Thermodynamics Overview: The study covering heat interactions, conversions to work, and four key laws governing thermodynamic systems:

    • Zeroth Law: Establishes thermal equilibrium principles.
    • First Law: Energy conservation perspective relating heat and work.
    • Second Law: Defines the unidirectional flow of heat, inciting order to disorder tendencies in mechanical energy.
    • Third Law: Proclaims entropy tendency toward constant values nearing absolute zero, building the quantum discussion scope in thermodynamic contexts.