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., \[\eta =\frac{Work\,done}{Heat\,\sup plied}=\frac{W}{{{Q}_{1}}}=\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}\] \[=1-\frac{{{Q}_{2}}}{{{Q}_{1}}}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] 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 \[{{T}_{2}}=300\,\,K\] 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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