Thermodynamics Question 284
Question: A Carnot engine takes $ 3\times 10^{6}cal. $ of heat from a reservoir at $ 627{}^\circ C, $ and gives it to a sink at $ 27{}^\circ C. $ The work done by the engine is
Options:
A) $ 4.2\times 10^{6}J $
B) $ 8.4\times 10^{6}J $
C) $ 16.8\times 10^{6}J $
D) zero
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ \eta =\frac{( 627 )+( 273 )-( 273+27 )}{627+273} $
$ =\frac{900-300}{900}=\frac{600}{900}=\frac{2}{3} $
$ \text{work=}( \eta )\times \text{Heat =}\frac{2}{3}\times 3\times 10^{6}\times 4.2J $
$ =8.4\times 10^{6}J $
