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Honors Physics Unit 11 Review: Thermal Energy, Heat, and Work

This unit is about what happens to energy when things heat up, cool down, melt, or boil. By the end, you should be able to tell temperature, thermal energy, and heat apart, and calculate how much energy it takes to change an object's temperature or phase.

Read the overview first, then work through each topic's notes and try the checkpoint question before checking the answer.

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What is Honors Physics unit 11?

Earlier units tracked kinetic and potential energy of whole objects. This unit zooms in to the particles inside an object and the energy associated with their motion and interactions. Energy conservation still runs the show, and it now includes heat.

Honors Physics Unit 11 covers thermal physics. You'll learn the difference between temperature and thermal energy, convert between Celsius, Fahrenheit, and Kelvin, and use Q = mcΔT with specific heat to find energy transfers and final temperatures in mixing (calorimetry) problems. You'll compare conduction, convection, and radiation. You'll also use Q = mL with latent heats of fusion and vaporization for phase changes at constant temperature. For a pure substance at constant pressure, temperature stays flat during melting or boiling. Many courses connect this to the first law of thermodynamics, ΔU = Q + W, where W is work done on the system.

Temperature and total thermal energy are different quantities.

A bathtub of 40°C water and a cup of 40°C water have the same temperature, but the bathtub has more thermal energy because it contains more water. For an ideal gas, temperature in kelvins is directly proportional to the average translational kinetic energy of its particles.

Heat is energy in transit, and it always flows from hot to cold on its own.

An object doesn't contain heat. It contains thermal energy, and heat is the name for the energy that crosses between objects at different temperatures. Heat can be transferred by conduction, convection, or radiation. At thermal equilibrium, there is no net heat transfer between the objects.

Heating curves can show both warming and phase changes.

For heating within a single phase, Q = mcΔT relates heat transfer to temperature change when specific heat is approximately constant and other energy transfers are negligible. For a pure substance melting or boiling at constant pressure, temperature stays constant while energy changes the arrangement and interactions of particles. Use Q = mL for that phase change. A heating curve shows these processes as sloped and flat segments.

Energy is conserved, even when it hides inside particles.

Every problem in this unit is an energy bookkeeping problem. If two objects exchange heat only with each other, heat lost by the hot object equals heat gained by the cold one. For a closed system, the first law is ΔU = Q + W, where Q is heat added and W is work done on the system. If you track where every joule goes, the formulas follow.

Honors Physics unit 11 topics

Unit 11 review notes

11.1

Temperature and Thermal Energy

Temperature describes how hot or cold a substance is. For an ideal gas, it is proportional to the average translational kinetic energy of the particles. Thermal energy includes energy associated with particle motion and interactions. A lake at 20°C has much more thermal energy than a cup of water at 100°C, even though the cup is hotter, because the lake has vastly more particles.

For an ideal gas, the average translational kinetic energy per particle is KEavg=32kBTKE_{avg} = \frac{3}{2}k_B T, where kB=1.38×10−23k_B = 1.38 \times 10^{-23} J/K and TT is in kelvins. Use kelvins because this formula requires absolute temperature. Doubling the Kelvin temperature doubles that average kinetic energy. Doubling a Celsius reading does not. Absolute zero is 0 K, although quantum systems can still have zero-point motion.

Use these conversions:

  • TK=TC+273.15T_K = T_C + 273.15
  • TF=95TC+32T_F = \frac{9}{5}T_C + 32
  • TC=59(TF−32)T_C = \frac{5}{9}(T_F - 32)

A kelvin and a Celsius degree are the same size, so a temperature change of 10°C is also a change of 10 K. A Fahrenheit degree is smaller, so a 10°C change is an 18°F change.

When two objects at different temperatures touch, heat flows from the hotter one to the colder one. If they exchange energy only with each other, they reach the same temperature and thermal equilibrium. The zeroth law says that if A and B are each in equilibrium with C (say, a thermometer), then A and B are in equilibrium with each other, which is why thermometers work.

  • Temperature: A measure of the average kinetic energy of the particles in a substance.
  • Thermal energy: Energy associated with particle motion and interactions in an object. It depends on temperature, the amount of material, and its state.
  • Absolute zero: 0 K, or -273.15°C, the lowest possible temperature and the zero point of the Kelvin scale.
  • Thermal equilibrium: The condition where objects in contact have the same temperature, so there is no net heat flow between them.

Check yourself: An ideal gas is heated from 27°C to 327°C. By what factor does the average translational kinetic energy of its particles increase? (Answer: Convert to kelvins first, giving 300 K and 600 K. The Kelvin temperature doubles, so the average kinetic energy doubles. It does not increase by a factor of about 12.)

QuantityWhat it describesDepends on amount of material?Units
TemperatureHow hot or cold a system isNoK, °C, °F
Thermal energyEnergy associated with particle motion and interactionsYesJ
HeatEnergy transferred because of a temperature differenceYesJ
11.2

Heat, Specific Heat, and Heat Transfer

Heat (QQ) is energy transferred because of a temperature difference. It's measured in joules, and 1 cal = 4.184 J. Specific heat (cc) is the energy needed to raise 1 kg of a material by 1°C. Water's is high, about 4186 J/(kg·°C), so it warms and cools slowly. Metals like iron (about 450 J/(kg·°C)) change temperature quickly.

For heating or cooling without a phase change, Q=mcΔTQ = mc\Delta T applies when specific heat is approximately constant and other energy transfers are negligible. A positive QQ means heat entered the object, and a negative QQ means heat left it.

Worked example. A 0.200 kg iron block at 90°C is dropped into 0.500 kg of water at 20°C. Find the final temperature TT. In an isolated system, heat lost by the iron equals heat gained by the water. Set up (0.200)(450)(90−T)=(0.500)(4186)(T−20)(0.200)(450)(90 - T) = (0.500)(4186)(T - 20). This becomes 8100−90T=2093T−41,8608100 - 90T = 2093T - 41{,}860. Rearranging gives 49,960=2183T49{,}960 = 2183T, so T≈22.9T \approx 22.9°C. The water barely warms because of its large mass and specific heat.

Heat moves in three ways. Conduction transfers energy through interactions between neighboring particles and, in metals, through mobile electrons. For steady conduction through a uniform slab, the rate is Qt=kAΔTd\frac{Q}{t} = kA\frac{\Delta T}{d}, where kk is conductivity, AA is area, and dd is thickness. Convection carries energy by moving fluid, as when hot air rises. Radiation sends energy as electromagnetic waves, even through a vacuum.

  • Specific heat capacity: The energy needed to raise the temperature of 1 kg of a substance by 1°C, measured in J/(kg·°C).
  • Calorimetry: Using energy transfer to find unknown quantities such as temperatures or specific heats. In an isolated system, total heat lost equals total heat gained.
  • Thermal conductivity: A material property, k, that describes how easily heat conducts through it, high for copper and low for foam.

Check yourself: How much heat does it take to warm 2.0 kg of water from 15°C to 65°C? (Answer: Q = mcΔT = (2.0)(4186)(50) = 418,600 J, or about 419 kJ.)

MethodHow energy movesNeeds a medium?Everyday example
ConductionParticle interactions and, in metals, mobile electronsYes, works best in solidsA metal spoon getting hot in soup
ConvectionBulk flow of a heated fluidYes, liquids and gases onlyWarm air rising from a heater
RadiationElectromagnetic wavesNo, works in a vacuumSunlight warming Earth
11.3

Phase Change and Latent Heat

When a pure substance melts or boils at constant pressure, its temperature stays constant during the phase change. The energy changes the arrangement and interactions of its particles. Latent heat LL is the energy transferred per unit mass. Use Q=mLQ = mL for heat absorbed and Q=−mLQ = -mL for heat released, with LL positive.

At about 1 atm, water's latent heat of fusion is Lf=3.34×105L_f = 3.34 \times 10^5 J/kg at 0°C, and its latent heat of vaporization is Lv=2.26×106L_v = 2.26 \times 10^6 J/kg at 100°C. Vaporization requires much more energy per kilogram than melting because particles become much more widely separated. Steam at 100°C can cause a more severe burn than water at 100°C because it releases energy when it condenses on skin.

Melting and vaporization absorb energy. Freezing and condensation release the same amount for the same mass at the same phase-change conditions.

Worked example. How much energy turns 0.100 kg of ice at -10°C into water at 20°C? Use cice=2100c_{ice} = 2100 J/(kg·°C). Step 1, warm the ice: Q1=(0.100)(2100)(10)=2100Q_1 = (0.100)(2100)(10) = 2100 J. Step 2, melt it: Q2=(0.100)(3.34×105)=33,400Q_2 = (0.100)(3.34 \times 10^5) = 33{,}400 J. Step 3, warm the water: Q3=(0.100)(4186)(20)=8372Q_3 = (0.100)(4186)(20) = 8372 J. The total is about 43,900 J. Melting takes most of the energy.

On a heating curve for a pure substance at constant pressure, sloped segments can use mcΔTmc\Delta T and phase-change plateaus use mLmL.

  • Latent heat of fusion: The energy per kilogram needed to melt a solid, or released when the liquid freezes, at constant temperature.
  • Latent heat of vaporization: The energy per kilogram needed to turn a liquid into a gas, or released when the gas condenses, at constant temperature.
  • Heating curve: A graph of temperature versus energy added. For a pure substance at constant pressure, it has flat segments during melting and boiling.

Check yourself: How much energy is released when 0.050 kg of steam at 100°C condenses into water at 100°C? (Answer: The energy released has magnitude mL_v = (0.050)(2.26 × 10^6) = 113,000 J, so Q = -113,000 J for the steam. The temperature stays at 100°C the whole time.)

Key terms

TermDefinition
TemperatureA measure of the average kinetic energy of the particles in a substance.
Thermal energyEnergy associated with particle motion and interactions in an object.
HeatEnergy transferred between objects because of a temperature difference, always flowing from hot to cold on its own.
Specific heat capacityThe energy needed to raise 1 kg of a material by 1°C, used in Q = mcΔT.
Latent heatEnergy per kilogram absorbed or released during a phase change at constant temperature, used in Q = mL.
ConductionHeat transfer through interactions between neighboring particles and, in metals, through mobile electrons.
ConvectionHeat transfer by the bulk movement of a heated liquid or gas.
RadiationHeat transfer by electromagnetic waves, which needs no medium and can cross a vacuum.
Kelvin scaleAn absolute temperature scale that starts at absolute zero and has degrees the same size as Celsius degrees.
First law of thermodynamicsFor a closed system, the rule that internal energy changes by the heat added plus the work done on the system, written ΔU = Q + W.

Mistakes to avoid in Thermal Energy, Heat, and Work

Treating temperature and heat as the same thing

Everyday language says something "has a lot of heat," which blurs the two ideas. In physics, temperature describes how hot or cold a system is, and heat is energy transferred because of a temperature difference. Ask yourself whether the problem is asking how hot something is or how much energy moved.

Using Celsius when a formula needs Kelvin

The ideal-gas formula KEavg=32kBTKE_{avg} = \frac{3}{2}k_B T requires absolute temperature in kelvins. For a temperature change ΔT in Q = mcΔT, Celsius and Kelvin give the same number, so either works there. Check what temperature scale a formula requires before substituting.

Using Q = mcΔT across a phase change

For a pure substance melting or boiling at constant pressure, ΔT is zero during the phase change, but energy is still transferred. Split the process into steps. Use mcΔT for each warming or cooling stretch and mL for each phase-change plateau, then add the pieces with their signs.

Mixing up signs in calorimetry

If you write ΔT as final minus initial for both objects, the hot object's Q comes out negative. If the two objects exchange energy only with each other, then Q_hot + Q_cold = 0. Alternatively, write heat lost = heat gained and use positive temperature differences on each side. Pick one method and stick with it. Mixing the two methods is how sign errors happen.

Study unit 11 other ways

Ready to review Unit 11?Start with the notes, check the topic cards, and use the practice or resource links when they are available for this course.