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 \[\eta =\frac{\Delta W}{\Delta {{Q}_{H}}}=1-\frac{{{T}_{c}}}{{{T}_{H}}}\] Given, \[{{T}_{c}}=27+273=300\,K\] \[{{T}_{H}}=727+273=1000\,K\] \[\eta =\left( 1-\frac{300}{1000} \right)\times 100=70%\]
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