First slide

Heat Engine; Carnot Engine and refrigerator

Question

Heat flows from a reservoir at 373 K to a reservoir at 273k through a copper rod as shown in the figure. The heat then leaves the 273K reservoir and enters a Carnot engine, which uses part of this heat to do work and rejects the remainder to a third reservoir at 173K.What fraction of the heat leaving the 373K reservoir is rendered unavailable for doing work, as compared to the situation where a Carnot engine is connected directly between the 373K and 173K reservoirs?

Moderate

A

3%
B

10%
C

17%
D

35%

Solution

[Concept]: With reservoirs at 273K,173K
[Step 1]: $\frac{W}{Q} = \frac{273 - 173}{273} = 0.37$
[Concept]: With reservoirs at 373K, 173K
[Step 2]: $\frac{W}{Q} = \frac{373 - 173}{373} = 0.54$
[Step 3]:
$\begin{array}{l} W^{'} - W = (0.54 - 0.37) Q = 0.17 Q \\ \Rightarrow \frac{W^{'} - W}{Q} = 0.17 \\ \Rightarrow (\frac{W^{'} - W}{Q}) 100 = 0.17 \times 100 = 17 % \end{array}$

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