A) 600 K
B) 500 K
C) 400 K
D) 100 K
Correct Answer: C
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 \[0.5=1-\frac{500}{{{T}_{1}}}\] 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 \[T_{2}^{\,'}=0.4\times 1000=400K\] 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.
You need to login to perform this action.
You will be redirected in
3 sec
