A) 0.75
B) 0.4
C) 0.25
D) 0.5
Correct Answer: D
Solution :
The efficiency \[(\eta )\] of Carnot engine is defined as the amount of work divided by the heat transferred between the system and the hot reservoir. i.e., \[\eta =\frac{\Delta \,W}{\Delta \,{{Q}_{H}}}=1-\frac{{{T}_{C}}}{{{T}_{H}}}\] Given, \[{{T}_{H}}={{27}^{o}}C=273+27=300\,K\], \[{{T}_{C}}=-{{123}^{o}}C=273-123=150\,K\] \[\therefore \] \[\eta =1-\frac{150}{300}=0.5\]
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