Question

Question

Entropy Analysis of the heat engine: consider a 35% efficient heat engine operating between a large, high-temperature reservoir at 1000 K (727 °C) and a large, cold reservoir at 308 K (35°C).
a. If it withdraws 1.2 MJ/s from the high-temperature reservoir, what would be the rate of loss of entropy from that reservoir and what would be the rate of gain by the low-temperature reservoir?
b. Express the work done by the engine in watts.
c. What would be the total entropy gain of the system?
d. Determine Carnot efficiency and recalculate the a, b, and c, accordingly.

Answer 1

a. The rate of loss of entropy from the high-temperature reservoir is 1.2 MJ/s x (1 - 0.35) = 0.78 MJ/s, and the rate of gain of entropy by the low-temperature reservoir is 1.2 MJ/s x 0.35 = 0.42 MJ/s.

b. The work done by the engine is 1.2 MJ/s x 0.35 = 0.42 MW.

c. The total entropy gain of the system is 0.42 MJ/s x (1000 K - 308 K) = 124.8 MJ/s.

d. = 83 MJ/s

Step-by-step explanation: