A Carnot engine has an efficiency of 25% and rejects heat to a sink at 27 °C. What is the temperature of its source?
- Sink temperature T2 = 27 + 273 = 300 K.
- 1 − T2/T1 = 0.25, so T2/T1 = 0.75.
- T1 = 300/0.75 = 400 K = 127 °C.
Answer: 400 K (127 °C)
Kinetic theory and thermodynamics
NUST NET Physics · Physics · Heat and thermodynamics. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
The idea
A gas is a huge number of molecules in random motion. Their collisions with the walls cause the pressure, and the absolute temperature measures their average translational kinetic energy. So every gas formula needs temperature in kelvin (K = °C + 273).
The first law of thermodynamics is energy conservation: heat supplied to a gas either raises its internal energy or does work as the gas expands. For an ideal gas, internal energy depends on temperature only.
A heat engine takes heat from a hot source, turns part of it into work and rejects the rest to a cold sink. No engine can beat a Carnot engine working between the same two temperatures.
Rules to remember
- PV = nRT = NkT; R = 8.31 J mol−1 K−1, k = 1.38 × 10−23 J K−1
- Pressure P = (1/3)ρ⟨v2⟩; mean translational KE per molecule = (3/2)kT
- vrms = √(3RT/M) = √(3kT/m): proportional to √T and to 1/√(molar mass)
- First law: Q = ΔU + W, where W = PΔV is the work done by the gas at constant pressure
- Constant volume: W = 0, Q = ΔU. Isothermal: ΔU = 0, Q = W. Adiabatic: Q = 0, W = −ΔU
- Cp − Cv = R; γ = Cp/Cv; monatomic Cv = 3R/2, γ = 5/3; diatomic Cv = 5R/2, γ = 7/5
- Efficiency = W/Q1 = 1 − Q2/Q1; Carnot engine: 1 − T2/T1 (T1 source, T2 sink, in kelvin)
Common mistake
Leaving temperatures in °C. Going from 27 °C to 54 °C does not double the pressure, because in kelvin it is 300 K to 327 K. Convert before using any ratio.
Worked example
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