A) \[90\,{}^\circ C,\text{ }37\,{}^\circ C\]
B) \[99\,{}^\circ C,\text{ }37\,{}^\circ C\]
C) \[372\,{}^\circ C,\text{ }37\,{}^\circ C\]
D) \[206\,{}^\circ C,\text{ }37\,{}^\circ C\]
Correct Answer: B
Solution :
| Key Idea: The efficiency of heat engine is the ratio of work done to the heat taken from the source. |
| If \[{{T}_{1}}\] is temperature of source and \[{{T}_{2}}\] the temperature of sink, the efficiency of engine |
| \[\eta =\frac{\text{Work}\,\text{done}\,(W)}{\text{Heat}\,\text{taken}\,({{Q}_{1}})}\] |
| \[=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] |
| \[\therefore \] \[1-\frac{{{T}_{2}}}{{{T}_{1}}}=\frac{1}{6}\] ....(i) |
| When temperature of sink is reduced by \[62{{\,}^{o}}C\]then |
| \[T_{2}^{'}={{T}_{2}}-62\] |
| \[\therefore \] \[\eta '=1-\frac{T_{2}^{'}}{{{T}_{1}}}\] |
| Given \[\eta '=2\eta =\frac{2}{6}=\frac{1}{3}\] |
| \[\therefore \] \[\frac{1}{3}=1-\frac{{{T}_{2}}-62}{{{T}_{1}}}\] (ii) |
| From Eq. (i) |
| \[\frac{{{T}_{2}}-62}{{{T}_{1}}}=\frac{2}{3}\] ....(iv) |
| Dividing Eq. (iii) by Eq. (iv) |
| \[\frac{{{T}_{2}}}{{{T}_{2}}-62}=\frac{5}{4}\] |
| \[\Rightarrow \] \[4{{T}_{2}}=5{{T}_{2}}-310\] |
| \[\Rightarrow \] \[{{T}_{2}}=310\,K\] |
| and from Eq. (iii), we have |
| \[\frac{310}{{{T}_{1}}}=\frac{5}{6}\] |
| \[\Rightarrow \] \[{{T}_{1}}=372\,K\] |
| Hence, \[{{T}_{1}}=372\,K=372-273={{99}^{o}}C\] |
| and \[{{T}_{2}}=310\,K=310-273={{37}^{o}}C\] |
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