Thermodynamics Question 283
A refrigerator with coefficient of performance releases 200 J of heat to a hot reservoir. Then the work done on the working substance is 100 J
Options:
A) $ \frac{100}{3}J $
B) $ 100J $
C) $ \frac{200}{3}J $
D) $ 150J $
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Answer:
Correct Answer: D
Solution:
[d] The coefficient of performance of a refrigerator is given by $ \alpha =\frac{Q _{2}}{W}=\frac{Q _{2}}{Q _{1}-Q _{2}} $
Substituting the given values, we get $ \frac{1}{3}=\frac{Q _{2}}{200-Q _{2}} $
$ \Rightarrow 200-Q _{2}=3Q _{2}\Rightarrow 4Q _{2}=200 $
$ \text{or }Q _{2}=\frac{200}{4}J=50J $
$ \therefore W=Q _{1}-Q _{2}=200J-50J=150J $
BETA
