Thermodynamics · AP Physics 2 · 6 min
Thermodynamics problems become clearer once heat is treated as energy transfer, not a substance stored in an object.
01 The trap
Wrong path
Students compare temperatures to decide heat transfer, work, and internal energy without tracking the process.
Why it feels right
Everyday language says an object 'has heat,' so hotter can feel like more total thermal energy and heat added can feel like work output.
02 Correct model
Temperature measures average microscopic kinetic energy. Heat Q is energy transferred because of a temperature difference. Internal energy changes according to the first law, with signs fixed by the convention in the problem.
03 Mini-example
Prompt
An ideal gas expands isothermally. What is ΔU?
Common wrong answer
Positive because heat entered the gas.
Correct reasoning
For an ideal gas, internal energy depends only on temperature, so ΔU = 0 during an isothermal process.
Diagnostic cue
If a problem gives a PV process, do not start with formulas; first name what stays fixed and what area under the curve means.
04 Guided practice
Warm-up isolates the principle. Bridge changes the context. Contest-style requires a complete setup on less familiar geometry.
A small spark is at a much higher temperature than a bathtub of warm water. Which can contain more internal thermal energy?
Temperature is not total internal energy.
The bathtub can contain far more internal thermal energy because it has vastly more particles, even though its temperature is lower.
Trap: Equating higher temperature with more total thermal energy.
An ideal gas is heated at constant volume and absorbs 80 J of heat. What are W and ΔU using W as work done by the gas?
PV work is area under the curve; no volume change means no work.
W = 0 because ΔV = 0. With ΔU = Q - W, ΔU = 80 J.
Trap: Assuming any heat input must produce work.
An ideal gas is compressed adiabatically. Does its temperature increase, decrease, or stay constant?
During compression, work is done on the gas.
The temperature increases. With Q = 0 and compression meaning work done by the gas is negative, ΔU = Q - W is positive, so the ideal gas temperature rises.
Trap: Interpreting adiabatic as constant temperature.
A gas absorbs 250 J of heat and does 80 J of work on its surroundings. What is the change in its internal energy?
Use ΔU = Q - W.
ΔU = 250 J - 80 J = +170 J.
Trap: Adding the work even though energy leaves the gas through work on the surroundings.