Heat and Temperature: Comprehensive Study Notes

Heat: a form of energy in transit between bodies of different temperatures

• Distinction from internal energy: heat flows between bodies; internal energy is the state energy of a system
• Heat can increase kinetic energy (molecular motion) or potential energy (position) within a substance (e.g., melting, evaporation)
• All other forms of energy can be converted to heat (and heat to other forms) example: mechanical work (friction or compression) converts to heat
• When heat is transmitted to/from a system, the internal energy of the system changes
• Conceptual idea: heat is energy in transit, not a substance stored inside the body

Distinction: Heat vs Temperature vs Internal Energy

• Internal Energy (thermal): total microscopic energy of a system due to molecular motion and interactions
• Temperature: a measure of the average kinetic energy of molecules; indicates the direction of heat flow between objects in contact
• Heat: energy transferred due to a temperature difference; measured by the energy transferred, not by the amount of substance
• Thermometer: measures temperature; based on expansion/contraction of materials with temperature

Temperature Scales and Conversions

• Boiling point of pure water: 100° on the centigrade (C) scale; 212° on the Fahrenheit (F) scale
• Freezing point of pure water: 0° C; 32° F
• Centigrade (Celsius) scale: 0 to 100 between freezing and boiling points; divided into 100 equal degrees
• Fahrenheit (F) scale: between freezing and boiling points, divided into 180 equal parts
• Absolute zero: the lowest possible temperature; all molecular motion ceases
• On the Celsius scale:
  -273.15° C is absolute zero
• On the Kelvin scale: 0 K corresponds to −273.15°C

Temperature Conversion Formulas (all in LaTeX)

• Celsius to Fahrenheit: $^ ext{o}F = \frac{9}{5} imes {}^ ext{o}C + 32$
• Fahrenheit to Celsius: $^ ext{o}C = \frac{5}{9} imes ({}^ ext{o}F - 32)$
• Celsius to Kelvin: $K = {}^ ext{o}C + 273.15$
• Kelvin to Celsius: ${}^ ext{o}C = K - 273.15$
• Absolute zero reference: $0~ ext{K} ext{ corresponds to } -273.15^ ext{o}C$

Worked Conversion Examples (results shown; steps follow formulas)

• Convert 80 °F to °C: $^ ext{o}C = \frac{5}{9}(80 - 32) = \frac{5}{9} imes 48 = 26.7^ ext{o}C$
• Convert 80 °C to °F: $^ ext{o}F = \frac{9}{5} imes 80 + 32 = 176^ ext{o}F$
• Classroom temperature 24.0 °C to Kelvin: $K = 24.0 + 273.15 = 297.15 ext{ K} o 297 ext{ K (rounded)}$
• Liquid nitrogen at 77.0 K to °C: $^ ext{o}C = 77.0 - 273.15 = -196.15^ ext{o}C$
• Boiling point of liquid oxygen in °C to K: if T = -183 °C, then $K = -183 + 273.15 ext{ K} \, o 90.15 ext{ K}$
• Oxygen freezing at -362 °F to °C: use $^ ext{o}C = \frac{5}{9}({}^ ext{o}F-32)$ to obtain approximately $-219^ ext{o}C$ for the given value

Summary of Temperature-Change Problems (concepts)

• A change of 20 °C corresponds to a Fahrenheit change of 68 °F (using F = (9/5)C + 32)
• A temperature change of 95 °F corresponds to a Celsius change of about 35 °C (ΔC = ΔF×5/9 with ΔF = 95 − 32 or as appropriate for the problem)
• A temperature drop of 27 °F corresponds to a Celsius drop of 15 °C (ΔC = ΔF × 5/9 with ΔF = −27)
• Conversions between scales preserve the physical meaning of a temperature difference via the 5/9 factor

Heat and Heat Measurement

• Symbol: $Q$ denotes heat transferred
• Law of Conservation of Energy: energy cannot be created or destroyed; it can be transformed from one form to another
• Mass–energy equivalence: $E = m c^2$; combines conservation of energy and mass into a broader mass–energy conservation principle
• SI unit for heat: Joule (J)
• Other common units: calorie (cal), kilocalorie (kcal), British thermal unit (Btu)

Unit Conversions and Common Equivalents

• 1 J = 0.2390 cal
• 1 cal = 4.186 J (commonly rounded to 4.184 or 4.186 J)
• 1 kcal = 4186 J = 1000 cal
• 1 Btu = 1055 J
• 1 kcal = 3.969 Btu
• 1 Btu = 252 cal ≈ 0.252 kcal
• Equivalent forms: 1 cal = 4.186 J; 1 kcal = 4186 J; 1 Btu ≈ 1055 J

Specific Heats and Heat Capacity

• Specific heat (c): amount of heat required to raise the temperature of 1 g of a substance by 1°C; SI unit: J/(kg·°C) or cal/(g·°C)
• Water has a high specific heat: $c_{ ext{water}} = 4.186 ext{ J/(g·°C)} = 4186 ext{ J/(kg·°C)} = 1.00 ext{ cal/(g·°C)}$
• Ice: $c_{ ext{ice}} \, ext{≈} \, 2.090 ext{ kJ/(kg·°C)} = 0.50 ext{ kcal/(kg·°C)} \, (≈ 2.090 ext{ kJ/kg°C})$
• Steam (vapor): $c_{ ext{steam}} \, ext{≈} \, 2.010 ext{ kJ/(kg·°C)} \, (≈ 0.46 ext{ kcal/(kg·°C)})$
• Latent heat concepts (no temperature change during phase change):
• Heat of fusion (L_f): energy to melt 1 g of a solid at its melting point without a temperature change
  • Ice: $L_f ext{(ice → water)} = 80 ext{ cal/g} = 333 ext{ J/g} = 2.56 ext{ kcal/kg}$
  • English units example: 144 Btu/lb for ice fusion at 32°F
• Heat of vaporization (L_v): energy to vaporize 1 g of a liquid at its boiling point without a temperature change
  • Water: $L_v ext{(water → steam)} \,= 540 ext{ cal/g} = 2256 ext{ J/g} = 2.256 ext{ kJ/g}$
• General heat equation when temperature changes (for a given mass and specific heat):
• $Q = m c \, ext{Δ}T$
• For phase changes (no ΔT):
• Melting: $Q = m L_f$
• Vaporization: $Q = m L_v$
• Thermal energy balance examples (calorimetry) illustrate how to compute heat gained or lost by different substances

Change of State (Phase Transitions)

• Transitions: solid → liquid (melting), liquid → gas (evaporation/boiling), gas → liquid (condensation)
• During phase change, temperature remains constant while energy goes into breaking/interacting molecular bonds
• Melting point and freezing point occur at a specific temperature; pressure can affect melting point for substances that expand on freezing
• Latent heat is the energy exchanged without a temperature change during phase transitions
• Evaporation vs boiling
• Evaporation: occurs at the surface at any temperature
• Boiling: occurs throughout the liquid at a characteristic boiling temperature
• Effects of environment on phase change: presence of dissolved substances lowers freezing point; pressure effects depend on substance
• Example problem types include calorimetry involving ice melting and mixtures of water and ice, or vaporization calculations

Heat and Expansion

• Heating typically causes expansion in solids, liquids, and gases; extent depends on:
• Material type (solid, liquid, gas)
• Original size/shape
• Temperature change magnitude
• Coefficients of expansion:
• Linear expansion: ext{Δ}L = oldsymbol{
  } L_0 \Delta T with the coefficient of linear expansion
  (often denoted α)
• Volume expansion: $ext{Δ}V = \beta V_0 \Delta T$ where β is the coefficient of volume expansion (≈ 3α for many solids)
• Area expansion: $ext{Δ}A = ext{γ} A_0 \Delta T$ where γ is the coefficient of area expansion
• Gases expand much more than liquids, which in turn expand more than most solids
• Examples (typical problems):
• Steel rail 50 ft long, ΔT = 25°C, α_steel ≈ 1.2×10^-5 /°C
  • ΔL ≈ 50 × 1.2×10^-5 × 25 = 0.015 ft
• Aluminum strip 10 ft long, ΔT = 40°C, α_Al ≈ 2.2×10^-5 /°C
  • ΔL ≈ 10 × 2.2×10^-5 × 40 = 0.0088 ft; new length ≈ 10.0088 ft
• Copper wire 300 ft long, ΔT = 60°C, α_Cu ≈ 1.7×10^-5 /°C
  • ΔL ≈ 300 × 1.7×10^-5 × 60 ≈ 0.306 ft; new length ≈ 299.694 ft
• Practical example: a wheel tire fit uses differential expansion; heating a tire to around 299°C may be needed to fit over a wheel with 119.6 cm inner diameter to an outer diameter of 120 cm, using α_steel ≈ 1.2×10^-5 /°C; solving ΔL = 0.4 cm yields ΔT ≈ 279°C; final temperature ≈ 299°C
• Volume expansion of vessels and liquids can cause overflow when heated; example calculation shows overflow volume given vessel and liquid expansion coefficients

Gas Laws and the Ideal Gas Law

• Boyle's Law (constant T): P V = constant; at constant temperature, increasing pressure reduces volume
• Charles' Law (constant P): V ∝ T (absolute temperature); equation form:
  $\frac{V<em>1}{T</em>1} = \frac{V<em>2}{T</em>2} ext{ (T in absolute units)}$
• Ideal Gas Law: $P V = n R T$
• Connects pressure, volume, amount of substance (n), and absolute temperature (T)
• Practical note: use absolute temperature (K) in gas-law calculations

Heat Transfer Modes

• Conduction: heat transfer by molecular collisions and transfer of kinetic energy; least effective in gases due to larger molecular spacing
• Example: heat conduction along a copper rod from hot end to cold end
• Convection: transfer in fluids (liquids and gases) via convection currents; warmer portions rise, colder portions sink
• Example: heating water in a boiler from below creates currents that distribute heat
• Radiation: heat transfer by electromagnetic waves; does not require a medium; can travel through space
• Examples: heat from the Sun reaching Earth; a blackbody absorbs/emits radiation efficiently

Solved Problems and Exercises (conceptual guidance)

• Calorimetry problems illustrate energy balance: heat lost by one substance equals heat gained by another
• Typical relations used:
• Heat gained by cold mass: $Q = m c \, \Delta T$
• Heat lost by hot mass: $Q = m c \Delta T$
• For mixtures of water and ice, account for melting heat, then sensible heating of resulting water
• Example problem outline: calorimeter with water and ice
• Determine heat exchanged by water, ice, and calorimeter, then compute latent heat of fusion per gram

Practical Summary of Key Formulas (in LaTeX)

• Temperature conversions:
• $^ ext{o}C = \frac{5}{9} ({}^ ext{o}F - 32)$
• $^ ext{o}F = \frac{9}{5} ({}^ ext{o}C) + 32$
• $K = {}^ ext{o}C + 273.15$
• Heat and phase changes:
• $Q = m c \, ext{Δ}T$
• $Q = m L_f ext{ (fusion)}$
• $Q = m L_v ext{ (vaporization)}$
• Density/expansion:
• ext{Δ}L =
  L_0 \, ext{Δ}T
• $ext{Δ}V = \beta V_0 \, ext{Δ}T$
• $ext{Δ}A = ext{γ} A_0 \, ext{Δ}T$
• Heat units relations:
• $1\text{ J} = 0.2390\text{ cal}$
• $1\text{ cal} = 4.186\text{ J}$
• $1\text{ kcal} = 4186\text{ J}$
• $1\text{ Btu} = 1055\text{ J}$
• $1\text{ kcal} = 3.969\text{ Btu}$
• Specific heats:
• water: $c_{ ext{water}} = 4.186\,\frac{\text{J}}{\text{g}~!\text{°C}} = 4186\,\frac{\text{J}}{\text{kg}~!\text{°C}}$
• ice: $c_{ ext{ice}} \approx 2.090\times 10^3\,\frac{\text{J}}{\text{kg}~!\text{°C}}$
• steam: $c_{ ext{steam}} \approx 2.010\times 10^3\,\frac{\text{J}}{\text{kg}~!\text{°C}}$
• latent heats (typical values):
• fusion of ice: $L_f = 80\,\frac{\text{cal}}{\text{g}} = 333\,\frac{\text{J}}{\text{g}}$
• vaporization of water: $L_v = 540\,\frac{\text{cal}}{\text{g}} = 2260\,\frac{\text{J}}{\text{g}}$

Quick Reference: why water moderates climate

• Water has high specific heat and high heat capacity per unit mass, allowing large bodies of water to absorb and store heat with only modest temperature changes
• This reservoir effect stabilizes coastal climates and seasonal temperature variations by absorbing heat in summer and releasing it in winter

Exercises and Practice Focus (topics in the provided material)

• Temperature scale conversions and absolute temperature usage in problems
• Determining final temperatures in multi-substance heat transfer problems using Q = mcΔT and Q = mLf or mLv as appropriate
• Using expansion coefficients to estimate length/volume changes with temperature
• Applying Boyle's Law, Charles' Law, and the Ideal Gas Law in context
• Conducting basic calorimetry experiments and interpreting results

Important Takeaways

• Heat is energy in transit; temperature measures the average molecular activity; internal energy is the state energy depending on structure and motion
• Phase changes involve latent heat; temperature does not change during fusion or vaporization while energy is absorbed or released
• Specific heat determines how much energy is needed to raise a mass’s temperature; water’s high specific heat makes it a key regulator of climate and many engineering systems
• Real-world applications include calorimetry, climate moderation by large water bodies, expansion of structures with temperature, and gas behavior in engines and weather systems