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)
NUST NET (Engineering) · Physics · Heat and thermodynamics. A short explanation of the idea, the rules to remember, the mistake to avoid, a worked example and practice questions with answers.
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.
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.
Answer: 400 K (127 °C)
Try each one, then open the answer.
Answer: B. vrms = √(3RT/M), so vH/vO = √(32/2) = √16 = 4. 16 forgets the square root; it is the average kinetic energies, not the speeds, that are equal.
Answer: A. η = 1 − T2/T1 = 1 − 300/500 = 0.40, so W = 0.40 × 1000 = 400 J. 600 J is the heat rejected to the sink, and 667 J divides by the sink temperature instead of the source temperature.
Answer: B. Work done by the gas W = PΔV = 2.0 × 105 × 1.5 × 10−3 = 300 J (1 L = 10−3 m3). ΔU = Q − W = 800 − 300 = 500 J. 1100 J adds the work instead of subtracting it.
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