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).You need to login to perform this action.
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