A) 378 K
B) 310 K
C) 404 K
D) 296 K
Correct Answer: A
Solution :
Work done during the cycle, W = 150 - 100 = 50 J Now, efficiency of engine\[\eta =\frac{W}{Q}=\frac{50}{150}=\frac{1}{3}=0.33\]From Carnot theorem \[0.33\le 1-\frac{{{T}_{2}}}{{{T}_{1}}}=1-\frac{250}{{{T}_{1}}}\] \[\Rightarrow \]\[0.33\le \frac{{{T}_{1}}-250}{{{T}_{1}}}\]\[\Rightarrow \]\[0.33{{T}_{1}}\le {{T}_{1}}-250\]\[\Rightarrow \] \[\Rightarrow \]\[250\le 1-0.33\]\[\Rightarrow \]\[\frac{250}{0.66}\le {{T}_{1}}\Rightarrow 378.78\le {{T}_{1}}\] \[\therefore \]\[{{T}_{\min }}=378.1K\]
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