Thermodynamics Question 165
Question: A Carnot engine takes $ 3\times 10^{6}cal $ . of heat from a reservoir at 627°C, and gives it to a sink at 27°C. The work done by the engine is
[AIEEE 2003]
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:
$ \eta =1-\frac{T _{2}}{T _{1}}=\frac{W}{Q} $
Therefore $ W=( 1-\frac{T _{1}}{T _{2}} )\ Q={ 1-\frac{(273+27)}{(273+627)} } $
Therefore $ W=( 1-\frac{300}{900} )\times 3\times 10^{6} $
$ =2\times 10^{6}\times 4.2\ J=8.4\times 10^{6}J $
