A) 70%
B) 30%
C) 0%
D) 10%
Correct Answer: A
Solution :
The efficiency\[\eta \]of a Carnot engine is defined as the amount of work divided by the heat transferred between the system and the hot reservoir. \[\eta =\frac{\Delta W}{\Delta H}=1-\frac{{{T}_{C}}}{{{T}_{H}}}\] \[{{T}_{C}}=27+273=300\,K\] \[{{T}_{H}}=727+273=1000\,K\] \[\eta =\left( 1-\frac{300}{1000} \right)\times 100\] \[=70%\]
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