Heat and Energy Transfer — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Definitions of heat vs thermal energy, thermal equilibrium, three modes of heat transfer (conduction, convection, radiation), specific heat capacity, Stefan-Boltzmann law, and calculation methods for calorimetry and heat transfer problems tested on AP Physics 2.

You should already know: Temperature is a measure of average microscopic kinetic energy in a system. The first law of thermodynamics relates heat, work, and internal energy. Basic SI unit conversions for length and temperature.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Heat and Energy Transfer?

Heat is defined as the transfer of thermal energy between two systems that occurs exclusively due to a temperature difference between the systems. In AP Physics 2 notation, heat is denoted $Q$, with the sign convention that $Q>0$ means heat is added to the system of interest, and $Q<0$ means heat leaves the system. Per the AP Physics 2 Course and Exam Description (CED), this topic makes up roughly 30% of Unit 2: Thermodynamics, which corresponds to 7-9% of the total exam score. Heat and energy transfer concepts appear in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with first law of thermodynamics or ideal gas problems. A key distinction to anchor your understanding: heat is energy in transit, not energy stored in a system. Stored thermal energy is called internal energy; temperature is the macroscopic driving force for heat transfer. Net heat transfer between two systems in thermal contact stops when both reach thermal equilibrium, meaning they have the same temperature.

2. Heat Capacity and Calorimetry

The relationship between heat transferred to a substance and its resulting temperature change is governed by specific heat capacity $c$, defined as the amount of energy required to raise the temperature of 1 kilogram of the substance by 1 Kelvin (or 1 degree Celsius, since temperature changes are identical in both scales). The fundamental formula for heat transfer and temperature change is: $$Q=mc\Delta T$$ where $m$ is mass of the substance, and $\Delta T=T_{final}-T_{initial}$ is the temperature change. For an insulated system (no heat exchanged with the surroundings), total energy is conserved, so total heat gained by colder substances equals total heat lost by warmer substances: $Q_{gained}=Q_{lost}$. This relationship is the basis of calorimetry, the experimental method for measuring specific heat capacity. Intuitively, substances with high specific heat (like water) require large amounts of heat to change temperature, making them useful for thermal storage, while substances with low specific heat (like metals) change temperature quickly when heated or cooled.

Worked Example

A 0.25 kg aluminum block at 85 °C is dropped into 0.75 kg of water at 22 °C in a perfectly insulated container. Find the final equilibrium temperature of the mixture, given $c_{Al}=900{\text{ J kg}}^{-1}°C^{-1}$ and $c_{water}=4186{\text{ J kg}}^{-1}°C^{-1}$.

1. Apply conservation of energy for the insulated system: heat lost by aluminum equals heat gained by water. Write the equation with positive terms on both sides to avoid sign errors: $m_{Al}{c}_{Al}(T_{Al,initial}-T_f)=m_w{c}_w(T_f-T_{w,initial})$
2. Rearrange to isolate $T_f$: $m_{Al}{c}_{Al}{T}_{Al,initial}+m_w{c}_w{T}_{w,initial}=T_f(m_{Al}{c}_{Al}+m_w{c}_w)$
3. Substitute numerical values: Left side = $(0.25)(900)(85)+(0.75)(4186)(22)=19125+69069=88194$ Denominator = $(0.25)(900)+(0.75)(4186)=225+3139.5=3364.5$
4. Solve for $T_f$: $T_f=88194/3364.5\approx 26.2°C$

Exam tip: If your final equilibrium temperature is outside the range of the two initial temperatures, you mixed up the direction of heat transfer. Rewrite the equation so both sides are positive to fix the error.

3. Conduction and Convection

There are three distinct modes of heat transfer, two of which require a material medium: conduction and convection. Conduction is heat transfer via molecular collisions through a stationary material (no bulk motion of the material itself). For steady-state conduction through a uniform slab of material, Fourier's Law gives the rate of heat transfer (power, $P=Q/\Delta t$): $$P=\frac{kA\Delta T}{L}$$ where $k$ is thermal conductivity (high for metals, low for insulators like air or wool), $A$ is the cross-sectional area of the slab, $L$ is the thickness of the slab, and $\Delta T$ is the temperature difference across the slab. Convection is heat transfer via bulk motion of a fluid (liquid or gas). Natural convection occurs when warm fluid is less dense than cool fluid, causing it to rise and carry heat away from a warm surface; forced convection occurs when an external force (like a fan or pump) moves fluid across the surface, increasing heat transfer rate. AP Physics 2 does not require a detailed formula for convection, but you must be able to identify when convection is the dominant mode of heat transfer.

Worked Example

A single-pane house window has an area of 1.5 m², thickness of 3.0 mm, and thermal conductivity of $0.8{\text{ W m}}^{-1}{K}^{-1}$. If indoor temperature is 20 °C and outdoor winter temperature is -5 °C, what is the rate of heat loss through the window?

1. Convert thickness to SI units: $L=3.0\text{ mm}=0.003\text{ m}$.
2. Calculate the temperature difference across the window: $\Delta T=20°C-(-5°C)=25K$ (temperature differences are identical in °C and K).
3. Substitute into Fourier's Law: $P=\frac{(0.8W{m}^{-1}{K}^{-1})(1.5m^2)(25K)}{0.003m}=10000W=10kW$

Exam tip: Always check the units of $k$ before plugging in length values. Most $k$ values use meters, so convert millimeters or centimeters to meters to avoid orders-of-magnitude errors.

4. Thermal Radiation

All objects with a temperature above absolute zero emit thermal radiation, which is electromagnetic radiation that can travel through a vacuum (no material medium required). The total power emitted by an object is given by the Stefan-Boltzmann Law: $$P=\sigma Ae{T}^4$$ where $\sigma =5.67\times 10^{-8}W{m}^{-2}{K}^{-4}$ is the Stefan-Boltzmann constant, $A$ is the surface area of the object, $e$ is emissivity (a value between 0 for a perfect reflector and 1 for a perfect blackbody radiator), and $T$ is the absolute temperature of the object in Kelvin. An object also absorbs radiation emitted by its surroundings, so the net rate of heat transfer via radiation is: $$P_{net}=\sigma Ae(T^4-T_s^4)$$ where $T_s$ is the absolute temperature of the surroundings. If $T>T_s$, the object has a net loss of heat via radiation; if $T<T_s$, it has a net gain. The $T^4$ dependence means radiation increases very rapidly with temperature.

Worked Example

A person has a body surface area of 1.6 m², emissivity of 0.7, and a skin temperature of 33 °C. If the room temperature is 20 °C, what is the net rate of heat loss from the person via radiation?

1. Convert temperatures to Kelvin: $T=33+273=306K$, $T_s=20+273=293K$.
2. Calculate the difference in $T^4$ terms: $T^4-T_s^4=(306{)}^4-(293{)}^4\approx 8.8\times 10^9-7.4\times 10^9=1.4\times 10^9{K}^4$.
3. Substitute into the net Stefan-Boltzmann equation: $P_{net}=(5.67\times 10^{-8}W{m}^{-2}{K}^{-4})(1.6m^2)(0.7)(1.4\times 10^9{K}^4)\approx 90W$

Exam tip: You must convert temperature to Kelvin for the Stefan-Boltzmann Law. Using Celsius will give a completely wrong result, because the law depends on absolute temperature, not temperature relative to freezing.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using Celsius temperature directly in the Stefan-Boltzmann law for thermal radiation. Why: Students are accustomed to using Celsius for $\Delta T$ in $Q=mc\Delta T$, so they forget the $T^4$ term requires absolute temperature. Correct move: Always add 273 to any temperature given in Celsius before plugging into the Stefan-Boltzmann equation.
• Wrong move: Mixing up heat and internal energy, claiming a warm object "contains a lot of heat". Why: Everyday language uses "heat" interchangeably with "hotness", which conflicts with the strict physics definition. Correct move: Always refer to heat as energy transferred; stored thermal energy in a system is called internal energy.
• Wrong move: Getting a final equilibrium temperature outside the range of the two initial temperatures in a calorimetry problem. Why: Students blindly plug $\Delta T=T_f-T_i$ into $Q=mc\Delta T$ for all substances, leading to incorrect sign cancellation. Correct move: Write energy conservation as $Q_{gained}=Q_{lost}$, so both sides of the equation are positive, eliminating sign errors.
• Wrong move: Claiming all heat transfer requires a material medium, and that no heat can transfer through vacuum. Why: Students confuse the requirements for conduction/convection with radiation. Correct move: Use the mnemonic: Radiation travels through space (vacuum), Conduction conducts through solids, Convection conveys heat via moving fluids.
• Wrong move: Leaving thickness in millimeters when calculating conduction power. Why: Insulation and window thicknesses are often quoted in millimeters for convenience, and students forget unit conversion. Correct move: Always match your length units to the units of thermal conductivity $k$, which almost always uses meters.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

A 1 kg block of copper at 100 °C is brought into thermal contact with a 4 kg block of copper at 0 °C in an insulated container. No work is done. What is the approximate final equilibrium temperature of the two blocks? A) 20 °C B) 50 °C C) 80 °C D) Cannot be determined without the specific heat of copper

Worked Solution: Use conservation of energy for insulated calorimetry: heat lost by the hot block equals heat gained by the cold block, so $m_{1}c(100-T_f)=m_{2}c(T_f-0)$. The specific heat $c$ is the same for both blocks, so it cancels out. Substituting masses gives $1(100-T_f)=4T_f$, which simplifies to $100=5T_f$, so $T_f=20°C$. The most common mistake is assuming temperature averages to 50 °C regardless of mass, which ignores the different total heat capacities of the two blocks. The correct answer is A.

Question 2 (Free Response)

A passive solar home uses a concrete slab floor to store heat from sunlight during the day. The slab has a mass of 5000 kg and specific heat capacity of $800Jk{g}^{-1}°C^{-1}$. (a) How much total heat energy does the slab store if its temperature increases from 18 °C in the morning to 28 °C in the late afternoon? (b) At night, the room and surrounding walls cool to 15 °C. The slab has a surface area of 15 m² and emissivity of 0.9. Calculate the net rate of heat loss from the slab when its temperature is 25 °C. (c) Explain why convection also contributes to heat loss at night, and how a ceiling fan would change the total rate of heat transfer.

Worked Solution: (a) Use the heat-temperature relationship $Q=mc\Delta T$. $\Delta T=28-18=10°C$. Substitute values: $Q=(5000\text{ kg})(800Jk{g}^{-1}°C^{-1})(10°C)=4\times 10^{7}J=40MJ$. (b) Convert temperatures to Kelvin: $T=25+273=298K$, $T_s=15+273=288K$. Use the net Stefan-Boltzmann equation: $P_{net}=\sigma Ae(T^4-T_s^4)=(5.67\times 10^{-8})(15)(0.9)(298^4-288^4)\approx 770W$. (c) The warm slab heats air in contact with it. Warm air is less dense than cool air, so it rises via natural convection, carrying heat away from the slab. A ceiling fan pushes additional cool air across the slab surface, increasing the rate of convective heat transfer compared to natural convection alone, leading to a higher total rate of heat loss.

Question 3 (Application / Real-World Style)

An adult polar bear has a 10 cm thick layer of blubber that insulates its core. The bear's core temperature is 37 °C, its outer skin temperature is 0 °C, total surface area covered by blubber is 2.0 m², and the thermal conductivity of blubber is $0.2W{m}^{-1}{K}^{-1}$. Calculate the rate of heat loss through the blubber, and interpret your result relative to the bear's typical metabolic heat production of ~300 W.

Worked Solution: First, convert blubber thickness to SI units: $L=10cm=0.1m$. The temperature difference across the blubber is $\Delta T=37-0=37K$. Apply Fourier's law for conduction: $P=\frac{kA\Delta T}{L}=\frac{(0.2W{m}^{-1}{K}^{-1})(2.0m^2)(37K)}{0.1m}=148W\approx 150W$. This rate of heat loss is about half of the polar bear's total metabolic heat production, meaning the blubber layer is effective at limiting heat loss enough to keep the polar bear warm in arctic conditions.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational prerequisite for the rest of thermodynamics in AP Physics 2. Next you will apply heat transfer concepts to the first and second laws of thermodynamics, where you will calculate work done by gases, changes in internal energy, and entropy changes for thermodynamic processes. Without mastering the distinction between heat and internal energy, and the ability to calculate heat transfer for different processes, you will not be able to correctly solve the first law problems that make up a large portion of Unit 2 exam questions. This topic also connects to electromagnetic thermal radiation in Unit 8, where you will explore the spectrum of blackbody radiation and its applications. Follow-on topics: First Law of Thermodynamics Entropy and the Second Law of Thermodynamics Thermal Properties of Ideal Gases Blackbody Radiation