AP Physics 2 Exam Question

A heat engine operates between two reservoirs at temperatures of 400 K and 800 K. During each cycle, 100 J of heat is absorbed from the high-temperature reservoir and 80 J of heat is exhausted to the low-temperature reservoir. Determine:

(a) The efficiency of this heat engine.

(b) The entropy change for the entire cycle.

Answer

(a) The efficiency of a heat engine can be calculated using the equation:

$\text{{Efficiency}} = \frac{{\text{{Work done by the engine}}}}{{\text{{Heat absorbed from the high-temperature reservoir}}}}$

In this case, the work done by the engine is given by:

$\text{{Work}} = \text{{Heat absorbed}} - \text{{Heat rejected}}$

Therefore, the efficiency is:

Substituting the given values:

So, the efficiency of this heat engine is 20%.

(b) The entropy change for the entire cycle can be determined using the equation:

$\Delta S = \frac{{Q_{\text{{rev}}}}}{{T}}$

where $\Delta S$ is the entropy change, $Q_{\text{{rev}}}$ is the heat absorbed or rejected in a reversible process, and $T$ is the temperature of the reservoir.

For the heat absorbed from the high-temperature reservoir, $Q_{\text{{rev}}} = 100 \, \text{J}$. The temperature of the high-temperature reservoir is $T = 400 \, \text{K}$.

Therefore, the entropy change for the heat absorbed is:

$\Delta S_{\text{{absorbed}}} = \frac{{Q_{\text{{rev}}}}}{{T}} = \frac{{100 \, \text{J}}}{{400 \, \text{K}}}$

For the heat rejected to the low-temperature reservoir, $Q_{\text{{rev}}} = 80 \, \text{J}$. The temperature of the low-temperature reservoir is $T = 800 \, \text{K}$.

Therefore, the entropy change for the heat rejected is:

$\Delta S_{\text{{rejected}}} = \frac{{Q_{\text{{rev}}}}}{{T}} = \frac{{80 \, \text{J}}}{{800 \, \text{K}}}$

Now, to calculate the total entropy change for the entire cycle, we add the entropy changes for the heat absorbed and rejected:

$\Delta S_{\text{{total}}} = \Delta S_{\text{{absorbed}}} + \Delta S_{\text{{rejected}}}$

$\Delta S_{\text{{total}}} = \frac{{100 \, \text{J}}}{{400 \, \text{K}}} + \frac{{80 \, \text{J}}}{{800 \, \text{K}}} = \frac{{1 \, \text{J}}}{{4 \, \text{K}}} + \frac{{1 \, \text{J}}}{{10 \, \text{K}}} = \frac{{5 + 2}}{20} = \frac{{7}}{20}$

Therefore, the entropy change for the entire cycle is $\frac{7}{20} \, \text{J/K}$.

Thus, the answers to the given questions are:

(a) The efficiency of this heat engine is 20%.

(b) The entropy change for the entire cycle is $\frac{7}{20} \, \text{J/K}$.