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 \[T{{}_{2}}=0.4\times 100=400\,K\] Note: Carnot engine is not a practical engine because many ideal situations have been assumed while designing this engine which can practically not be obtained.
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