Heat and Energy Transfer

AP Physics 2· AP Physics 2 CED — Thermodynamics· 14 min read

1. Core Definitions: Heat, Temperature, and Equilibrium★★☆☆☆⏱ 3 min

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. A key distinction: 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 stops when systems reach thermal equilibrium, meaning they have the same temperature. This topic makes up ~30% of Unit 2, or 7-9% of your total exam score, and appears in both MCQ and FRQ sections.

📘 Definition

Heat

$Q$

Energy transferred between systems due only to a temperature difference; heat is always energy in motion, not stored energy

Example:

Heat flows from a hot coffee mug to cool surrounding air

2. Heat Capacity and Calorimetry★★★☆☆⏱ 5 min

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 is:

Q = mc\Delta T

where $m$ is mass of the substance, and $Δ T = T_{f ina l} - T_{ini t ia l}$ 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: $Q_{g ain e d} = Q_{l os t}$ . This relationship is the basis of calorimetry, the experimental method for measuring specific heat capacity.

📐 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 $c_{Al} = 900 \text{ J kg}^{-1} ^\circ\text{C}^{-1}$ and $c_{water} = 4186 \text{ J kg}^{-1} ^\circ\text{C}^{-1}$ .

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:
2
$m_{Al}c_{Al}(T_{Al, initial} - T_f) = m_w c_w (T_f - T_{w, initial})$
3
Rearrange to isolate $T_{f}$ :
4
$m_{Al}c_{Al}T_{Al, initial} + m_w c_w T_{w, initial} = T_f (m_{Al}c_{Al} + m_w c_w)$
5
Substitute numerical values:
6
$\text{Left side} = (0.25)(900)(85) + (0.75)(4186)(22) = 19125 + 69069 = 88194 \\ \text{Denominator} = (0.25)(900) + (0.75)(4186) = 225 + 3139.5 = 3364.5$
7
Solve for $T_{f}$ :
8
$T_f = \frac{88194}{3364.5} \approx 26.2 ^\circ\text{C}$

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★★★☆☆⏱ 4 min

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, $P = Q /Δ t$ ):

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 $Δ T$ 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, only identification of when it is the dominant mode.

📐 Worked Example

A single-pane house window has an area of 1.5 m², thickness of 3.0 mm, and thermal conductivity of $0.8 W m^{- 1} K^{- 1}$ . 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:
2
$L = 3.0 \text{ mm} = 0.003 \text{ m}$
3
Calculate the temperature difference across the window (temperature differences are identical in °C and K):
4
$\Delta T = 20 ^\circ\text{C} - (-5 ^\circ\text{C}) = 25 K$
5
Substitute into Fourier's Law:
6
$P = \frac{(0.8 W m^{-1} K^{-1})(1.5 m^2)(25 K)}{0.003 m} = 10000 W = 10 kW$

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 and Stefan-Boltzmann Law★★★★☆⏱ 4 min

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 A e T^4

where $σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$ 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 from its surroundings, so the net rate of heat transfer via radiation is:

P_{net} = \sigma A e (T^4 - T_s^4)

where $T_{s}$ is the absolute temperature of the surroundings. If $T > T_{s}$ , the object has a net heat loss; 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 (required for the Stefan-Boltzmann law):
2
$T = 33 + 273 = 306 K, \quad T_s = 20 + 273 = 293 K$
3
Calculate the difference of $T^{4}$ terms:
4
$T^4 - T_s^4 = (306)^4 - (293)^4 \approx 8.8 \times 10^9 - 7.4 \times 10^9 = 1.4 \times 10^9 K^4$
5
Substitute into the net Stefan-Boltzmann equation:
6
$P_{net} = (5.67 \times 10^{-8} W m^{-2} K^{-4})(1.6 m^2)(0.7)(1.4 \times 10^9 K^4) \approx 90 W$

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

Wrong move:

Using Celsius temperature directly in the Stefan-Boltzmann law for thermal radiation

Why:

Students are accustomed to using Celsius for $Δ T$ in $Q = m c Δ T$ , 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 $Δ T = T_{f} - T_{i}$ into $Q = m c Δ T$ for all substances, leading to incorrect sign cancellation

Correct move:

Write energy conservation as $Q_{g ain e d} = Q_{l os t}$ , 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:

Remember: Radiation travels through vacuum, Conduction moves through stationary solids, Convection carries heat via moving fluids

Wrong move:

Leaving thickness in millimeters or centimeters when calculating conduction power

Why:

Thicknesses are often quoted in smaller units for convenience, and students forget unit conversion to match thermal conductivity units

Correct move:

Always match your length units to the units of thermal conductivity $k$ , which almost always uses meters

6. Quick Reference Cheatsheet

Category	Formula	Notes
Heat and temperature change	$Q = m c Δ T$	$Δ T$ can be in °C or K; $Q > 0$ = heat added, $Q < 0$ = heat lost
Calorimetry (insulated system)	$Q_{g ain e d} = Q_{l os t}$	Net heat transfer is zero; avoids sign errors
Steady-state conduction	$P = \frac{k A Δ T}{L}$	$k$ = thermal conductivity; $L$ = thickness; use meters for length
Net thermal radiation	$P_{n e t} = σ A e (T^{4} - T_{s}^{4})$	$T, T_{s}$ must be in Kelvin; $e$ = 0 (reflector) to 1 (blackbody); $σ = 5.67 \times 1 0^{- 8} W m^{- 2} K^{- 4}$
Convection	No formula required	Heat transfer via bulk fluid motion; requires a material medium
Thermal equilibrium	$T_{1} = T_{2}$	Net heat transfer between systems in contact is zero
Heat definition	$Q$	Energy transferred due to temperature difference; heat is energy in transit, not stored energy

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Calorimetry equilibrium problem
2022 · FRQ
Radiation heat loss calculation

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 real-world applications.