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. |
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