A) 380 K
B) 275 K
C) 325 K
D) 250 K
Correct Answer: D
Solution :
Efficiency of a Carnot engine, \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\]where, temperature of a source\[={{T}_{1}}\] Temperature of a sink\[={{T}_{2}}\] \[\therefore \] \[\frac{{{T}_{2}}}{{{T}_{1}}}=1-\eta =1-\frac{40}{100}=\frac{3}{5}\] Or \[{{T}_{1}}=\frac{5}{3}{{T}_{2}}=\frac{5}{3}\times 300=500K\] Increase in efficiency = 50% of 40% = 20% New efficiency \[\eta =40%+20%=60%\] \[\therefore \] \[\frac{{{T}_{2}}}{T_{1}^{'}}=1-\eta '=1-\frac{60}{100}=\frac{2}{5}\] \[T_{1}^{,}=\frac{5}{2}{{T}_{2}}=\frac{5}{2}\times 300=750\,K\] Increase in temperature of source \[=T_{1}^{'}-{{T}_{1}}\] \[=750-500=250\text{ }K\]
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