Question

During the isothermal heat rejection process of a Carnot cycle, the working fluid experiences an entropy change of -0.7 Btu/R. If the temperature of the heat sink is \(95^{\circ} \mathrm{F}\), determine \((a)\) the amount of heat transfer, \((b)\) the entropy change of the sink, and ( \(c\) ) the total entropy change for this process.

Short answer

Question: In an isothermal heat rejection process of a Carnot Cycle, the entropy change of the working fluid is -0.7 Btu/R, and the temperature of the heat sink is 95°F. Determine (a) the amount of heat transfer during the process, (b) the entropy change of the sink, and (c) the total entropy change for this process. Answer: (a) The amount of heat transfer during the isothermal heat rejection process is -388.27 Btu. (b) The entropy change of the sink is 0.7 Btu/R. (c) The total entropy change for this process is 0 Btu/R.

Step-by-step solution

Convert temperature to Rankine

Before starting with the calculations, we need to convert the temperature of the heat sink from Fahrenheit to Rankine. We can do this using the following formula: \(T_{\text{R}} = T_{\text{F}} + 459.67\) \(T_{\text{R}} = 95^{\circ} \mathrm{F} + 459.67 \Rightarrow T_{\text{R}} = 554.67 \,\text{R}\) Now, we have the temperature of the heat sink in Rankine.

Calculate the amount of heat transfer

Next, we will calculate the amount of heat transfer during the isothermal heat rejection process using the following equation: \(Q = T \Delta S\) Plugging in the values we have, the equation becomes: \(Q = 554.67 \,\text{R} \times (-0.7 \,\text{Btu/R})\) \(Q = -388.27 \,\text{Btu}\) The amount of heat transfer during the isothermal heat rejection process is -388.27 Btu.

Calculate the entropy change of the sink

We can calculate the entropy change of the sink using the following equation: \(\Delta S_{\text{sink}} = -\Delta S_{\text{fluid}}\) \(\Delta S_{\text{sink}} = -(-0.7 \,\text{Btu/R})\) \(\Delta S_{\text{sink}} = 0.7 \,\text{Btu/R}\) The entropy change of the sink is 0.7 Btu/R.

Calculate the total entropy change for this process

Finally, we will calculate the total entropy change for this process using the following equation: \(\Delta S_{\text{total}} = \Delta S_{\text{fluid}} + \Delta S_{\text{sink}}\) \(\Delta S_{\text{total}} = -0.7 \,\text{Btu/R} + 0.7 \,\text{Btu/R}\) \(\Delta S_{\text{total}} = 0 \,\text{Btu/R}\) The total entropy change for this process is 0 Btu/R. To summarize the results: (a) The amount of heat transfer during the isothermal heat rejection process is -388.27 Btu. (b) The entropy change of the sink is 0.7 Btu/R. (c) The total entropy change for this process is 0 Btu/R.