Discussion
A reversible engine receives 75 kJ/s of energy from a reservoir at 750 K and does 12 kJ/s of work. The heat is rejected to two reservoirs at 650 K and 550 K. The rate of heat rejection (in kJ/s) to the reservoir at 650 K is
Answer: D
The heat rejected to the 650 K reservoir is 52 kJ/s.
Energy balance: the total heat rejected is \(75 - 12 = 63\) kJ/s. A reversible engine produces no entropy, so the entropy taken from the hot reservoir equals the entropy given to the two cold ones. With \(Q_{1}\) to 650 K and \(Q_{2} = 63 - Q_{1}\) to 550 K:
\[ \begin{aligned} \frac{75}{750} &= \frac{Q_{1}}{650} + \frac{63 - Q_{1}}{550} \\ 0.1 - 0.11455 &= Q_{1}\left(\frac{1}{650} - \frac{1}{550}\right) \\ Q_{1} &= \frac{-0.01455}{-2.797\times10^{-4}} = 52.0 \end{aligned} \]
So 52 kJ/s goes to 650 K and 11 kJ/s to 550 K. The option 11 is the heat to the other reservoir; 31 and 41 do not satisfy the entropy balance.
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