A) \[600\,\,K\]
B) \[400\,\,K\]
C) \[500\,\,K\]
D) \[100\,\,K\]
Correct Answer: B
Solution :
Efficiency of the Carnot engine is given by \[\eta -1=\frac{{{T}_{2}}}{{{T}_{1}}}\] ... (i) where \[{{T}_{1}}=\]temperature of source \[{{T}_{2}}=\]temperature of sink Given,\[\eta =50%=0.5,\,\,{{T}_{2}}=500\,\,K\] Substituting in relation (i), we have or \[0.5=1-\frac{500}{{{T}_{1}}}\] or \[\frac{500}{{{T}_{1}}}=0.5\] \[\therefore \] \[{{T}_{1}}=\frac{500}{0.5}=1000\,\,K\] Now, the temperature of sink is changed to\[T{{'}_{2}}\] and the efficiency becomes\[60%\,\,i.e.,\,\,0.6\]. Using relation (i), we get \[0.6=1-\frac{T{{'}_{2}}}{1000}\] or \[\frac{T{{'}_{2}}}{1000}=1-0.6=0.4\] or \[T{{'}_{2}}=0.4\times 1000=400\,\,K\] Note: Carnot engine is not a practical engine because many idea] situations have been assumed while designing this engine which can practically not be obtained.
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