A) 275 K
B) 325 K
C) 250 K
D) 380 K
Correct Answer: C
Solution :
| The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied i.e., |
| \[\eta =\frac{\text{Work}\,\text{done}}{\text{Heat}\,\text{supplied}}=\frac{W}{{{Q}_{1}}}=\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}\] |
| \[=1-\frac{{{Q}_{2}}}{{{Q}_{1}}}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] |
| Here, \[{{T}_{1}}\] is the temperature of source and \[{{T}_{2}}\] is the temperature of sink |
| As given, \[\eta =40%=\frac{40}{100}=0.4\] |
| and \[{{T}_{2}}=300\text{ }K\] |
| So, \[0.4=1-\frac{300}{{{T}_{1}}}\] |
| \[\Rightarrow \] \[{{T}_{1}}=\frac{300}{1-0.4}=\frac{300}{0.6}=500\,K\] |
| Let temperature of the source be increased by x K, then efficiency becomes |
| \[\eta '=40%+50%\] of \[\eta \] |
| \[=\frac{40}{100}+\frac{50}{100}\times 0.4\] |
| \[=0.4+0.5\times 0.4\] |
| \[=0.6\] |
| Hence, \[0.6=1-\frac{300}{500+x}\] |
| \[\Rightarrow \] \[\frac{300}{500+x}=0.4\] |
| \[\Rightarrow \] \[500+x=\frac{3900}{0.4}=750\] |
| \[\therefore \] \[x=750-500=250\,K\] |
| Note: All reversible heat engines working between same temperatures are equally efficient and no heat engine can be more efficient than Carnot engine (as it is ideal). |
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