Thermodynamics — Hard Practice Quiz
A Physics cheat sheet for Thermodynamics — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.
Formulas & key concepts
First Law of Thermodynamics: Change in internal energy \(\Delta U\) equals heat \(Q\) added minus work \(W\) done by the system.
Heat Engine Efficiency: Efficiency \(e\) is work \(W\) divided by input heat \(Q_H\). Also related to rejected heat \(Q_C\).
Carnot Engine Efficiency: Maximum theoretical efficiency depends only on temperatures of hot \(T_H\) and cold \(T_C\) reservoirs.
Entropy Change: Change in entropy \(\Delta S\) for a reversible process at constant temperature \(T\).
Refrigerator Coefficient of Performance: Ratio of heat removed \(Q_C\) to work input \(W\).
Heat Pump Coefficient of Performance: Ratio of heat delivered \(Q_H\) to work input \(W\).
Practice quiz
A heat engine operates between a hot reservoir at $T_H = 600 \text{ K}$ and a cold reservoir at $T_C = 300 \text{ K}$. In each cycle, it absorbs $Q_H = 1200 \text{ J}$ from the hot reservoir. If its actual efficiency is $75 \%$ of the Carnot efficiency, how much work does the engine perform per cycle, and what is the change in internal energy of the working substance over one complete cycle?
- A) $W = 450 \text{ J}$, $\Delta U = 0 \text{ J}$
- B) $W = 600 \text{ J}$, $\Delta U = 0 \text{ J}$
- C) $W = 450 \text{ J}$, $\Delta U = 150 \text{ J}$
- D) $W = 900 \text{ J}$, $\Delta U = 0 \text{ J}$
Answer: A) $W = 450 \text{ J}$, $\Delta U = 0 \text{ J}$
A refrigerator removes $Q_C = 500 \text{ J}$ of heat from its cold compartment for every $150 \text{ J}$ of work input. If this same device were operated as a heat pump, how much heat would it deliver to the hot reservoir for $200 \text{ J}$ of work input, assuming its COP remains constant?
- A) $Q_H \approx 867 \text{ J}$
- B) $Q_H \approx 750 \text{ J}$
- C) $Q_H \approx 900 \text{ J}$
- D) $Q_H \approx 650 \text{ J}$
Answer: A) $Q_H \approx 867 \text{ J}$
An ideal gas undergoes a reversible isothermal expansion at $T = 400 \text{ K}$, performing $W = 2000 \text{ J}$ of work. What is the change in entropy of the gas during this process, and what is the heat absorbed by the gas?
- A) $\Delta S = 5 \text{ J/K}$, $Q = 2000 \text{ J}$
- B) $\Delta S = -5 \text{ J/K}$, $Q = 2000 \text{ J}$
- C) $\Delta S = 0 \text{ J/K}$, $Q = 0 \text{ J}$
- D) $\Delta S = 5 \text{ J/K}$, $Q = 0 \text{ J}$
Answer: A) $\Delta S = 5 \text{ J/K}$, $Q = 2000 \text{ J}$
A Carnot engine operates between two reservoirs. It absorbs $Q_H$ from the hot reservoir and rejects $Q_C$ to the cold reservoir. If the hot reservoir temperature is $T_H$ and the cold reservoir temperature is $T_C$, derive an expression for the work done by the engine in terms of $Q_C$, $T_H$, and $T_C$.
- A) $W = Q_C (\frac{T_H}{T_C} - 1)$
- B) $W = Q_C (1 - \frac{T_C}{T_H})$
- C) $W = Q_C \frac{T_H}{T_C}$
- D) $W = Q_C (\frac{T_C}{T_H} - 1)$
Answer: A) $W = Q_C (\frac{T_H}{T_C} - 1)$
A heat pump is used to heat a house, maintaining an indoor temperature of $T_H = 295 \text{ K}$ when the outdoor temperature is $T_C = 278 \text{ K}$. If the heat pump's actual coefficient of performance is $60 \%$ of the ideal Carnot COP, and it delivers $Q_H = 1.5 \times 10^5 \text{ J}$ of heat to the house per cycle, how much work input is required per cycle?
- A) $W \approx 1.44 \times 10^4 \text{ J}$
- B) $W \approx 2.45 \times 10^4 \text{ J}$
- C) $W \approx 8.82 \times 10^3 \text{ J}$
- D) $W \approx 1.00 \times 10^4 \text{ J}$
Answer: A) $W \approx 1.44 \times 10^4 \text{ J}$
A refrigerator operates between a cold reservoir at $T_C = 270 \text{ K}$ and a hot reservoir at $T_H = 300 \text{ K}$. Its actual COP is $80 \%$ of the Carnot COP. If it removes $Q_C = 1000 \text{ J}$ from the cold reservoir, how much heat is rejected to the hot reservoir?
- A) $Q_H \approx 1139 \text{ J}$
- B) $Q_H \approx 1111 \text{ J}$
- C) $Q_H \approx 1083 \text{ J}$
- D) $Q_H \approx 1250 \text{ J}$
Answer: A) $Q_H \approx 1139 \text{ J}$
A system undergoes a cyclic process where it absorbs $Q_H$ from a hot reservoir and rejects $Q_C$ to a cold reservoir, performing work $W$. Which of the following statements is always true for any real (irreversible) heat engine operating between two reservoirs?
- A) $\Delta U = 0$ and $\frac{Q_C}{Q_H} > \frac{T_C}{T_H}$
- B) $\Delta U > 0$ and $\frac{Q_C}{Q_H} = \frac{T_C}{T_H}$
- C) $\Delta U = 0$ and $\frac{Q_C}{Q_H} < \frac{T_C}{T_H}$
- D) $\Delta U < 0$ and $\frac{Q_C}{Q_H} > \frac{T_C}{T_H}$
Answer: A) $\Delta U = 0$ and $\frac{Q_C}{Q_H} > \frac{T_C}{T_H}$
A Carnot engine operates with a hot reservoir at $T_H$. If the cold reservoir temperature $T_C$ is increased such that the Carnot efficiency is halved, how does the ratio $\frac{T_C}{T_H}$ change? Assume $T_H$ remains constant.
- A) The new ratio is $1 - \frac{1}{2} (1 - \frac{T_C}{T_H})$
- B) The new ratio is $2 \frac{T_C}{T_H}$
- C) The new ratio is $\frac{1}{2} \frac{T_C}{T_H}$
- D) The new ratio is $1 + \frac{1}{2} (1 - \frac{T_C}{T_H})$
Answer: A) The new ratio is $1 - \frac{1}{2} (1 - \frac{T_C}{T_H})$
A reversible heat engine has an efficiency of $e = 0.40$. If this engine is run in reverse as a refrigerator, what is its coefficient of performance ($COP_{ref}$)?
- A) $COP_{ref} = 1.5$
- B) $COP_{ref} = 0.67$
- C) $COP_{ref} = 2.5$
- D) $COP_{ref} = 0.4$
Answer: A) $COP_{ref} = 1.5$
A heat pump delivers $Q_H = 2000 \text{ J}$ of heat to a house at $T_H = 290 \text{ K}$ while consuming $W = 500 \text{ J}$ of electrical energy. Assuming the process is reversible, what is the change in entropy of the cold reservoir (outside environment) from which heat is absorbed, and what is the COP of the heat pump?
- A) $\Delta S_C \approx 6.90 \text{ J/K}$, $COP_{hp} = 4$
- B) $\Delta S_C \approx 5.17 \text{ J/K}$, $COP_{hp} = 4$
- C) $\Delta S_C \approx 6.90 \text{ J/K}$, $COP_{hp} = 3$
- D) $\Delta S_C \approx 4.31 \text{ J/K}$, $COP_{hp} = 3$
Answer: A) $\Delta S_C \approx 6.90 \text{ J/K}$, $COP_{hp} = 4$
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