AP Physics 2 Exam Question - Entropy and Heat Engines

A Carnot heat engine operates between two reservoirs, a high-temperature reservoir at a temperature $T_{h}$, and a low-temperature reservoir at a temperature $T_{l}$. The engine absorbs heat $Q_{h}$ from the high-temperature reservoir and rejects heat $Q_{l}$ to the low-temperature reservoir.

a) Derive an expression for the efficiency ($\eta$) of a Carnot heat engine in terms of the temperatures of the reservoirs.

b) For a hypothetical Carnot heat engine operating between a high-temperature reservoir at 800 K and a low-temperature reservoir at 300 K, determine the maximum possible efficiency of the engine.

c) If the engine absorbs 400 J of heat from the high-temperature reservoir, determine the following:

i) The heat (Ql) rejected to the low-temperature reservoir.

ii) The work done by the engine.

iii) The entropy change of the high-temperature reservoir.

iv) The entropy change of the low-temperature reservoir.

Useful information:

The efficiency of a heat engine is given by: $\eta = 1 - \frac{T_{l}}{T_{h}}$

The entropy change of a reservoir is given by: $\Delta S = \frac{Q}{T}$, where $Q$ is the heat added or removed from the reservoir, and $T$ is the temperature of the reservoir.

Answer and Explanation:

a) To derive the expression for the efficiency of a Carnot heat engine, we use the First Law of Thermodynamics: $\Delta U = Q - W$, where $\Delta U$ is the change in internal energy, $Q$ is the heat added, and $W$ is the work done by the engine.

For a Carnot heat engine, $\Delta U = 0$ since it goes through a cycle and returns to its initial state. Thus, $Q = W$.

The efficiency of a heat engine is defined as: $\eta = \frac{W}{Q_{h}}$

Substituting $W$ with $Q$ in the above equation, we get: $\eta = \frac{Q}{Q_{h}}$

Since $Q = Q_{h} - Q_{l}$, we can rewrite the efficiency equation as: $\eta = \frac{Q_{h} - Q_{l}}{Q_{h}}$

b) Given the high-temperature reservoir temperature ($T_{h}$) = 800 K and the low-temperature reservoir temperature ($T_{l}$) = 300 K, we can use the efficiency equation to determine the maximum efficiency: $\eta = 1 - \frac{T_{l}}{T_{h}}$

Substituting the given values, we have: $\eta = 1 - \frac{300}{800} = 1 - 0.375 = 0.625$ or 62.5%

Therefore, the maximum possible efficiency of the hypothetical Carnot heat engine is 62.5%.

c) Given that the engine absorbs 400 J of heat ($Q_{h}$), we can determine the following:

i) To find the heat ($Q_{l}$) rejected to the low-temperature reservoir, we can use the heat conservation equation $Q = Q_{h} - Q_{l}$. Rearranging the equation, we get: $Q_{l} = Q_{h} - Q = 400 - 400 = 0$ J

Therefore, the heat rejected to the low-temperature reservoir is 0 J.

ii) The work done by the engine is equal to the heat absorbed ($Q_{h}$). Thus, the work done is 400 J.

iii) To determine the entropy change of the high-temperature reservoir, we use the entropy change formula: $\Delta S = \frac{Q}{T}$. Substituting the values, we have: $\Delta S = \frac{400}{800} = 0.5$ J/K

Therefore, the entropy change of the high-temperature reservoir is 0.5 J/K.

iv) Similarly, to find the entropy change of the low-temperature reservoir, we use the entropy change formula. Since the heat rejected ($Q_{l}$) is zero, the entropy change is also zero.

Therefore, the entropy change of the low-temperature reservoir is 0 J/K.

In summary: i) The heat rejected to the low-temperature reservoir is 0 J. ii) The work done by the engine is 400 J. iii) The entropy change of the high-temperature reservoir is 0.5 J/K. iv) The entropy change of the low-temperature reservoir is 0 J/K.