A) 275 K
B) 325 K
C) 250 K
D) 380 K
Correct Answer: C
Solution :
The efficiency of Cannot engine is defined as the ratio of work done to the heat supplied ie, \[\eta =\frac{work\text{ }done}{heat\text{ }supplied}=\frac{W}{{{Q}_{1}}}=\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}\] \[=1-\frac{{{Q}_{2}}}{{{Q}_{1}}}=1-\frac{{{T}_{2}}}{{{T}_{1}}}\] 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 \[{{T}_{2}}=300K\] So. \[0.4=1-\frac{300}{{{T}_{1}}}\] \[\Rightarrow \] \[{{T}_{1}}=\frac{300}{1-0.4}=\frac{300}{0.6}=500K\] Let temperature of the source be increased by x K, then efficiency becomes \[\eta =40%+50%\] of \[\eta \] \[=\frac{400}{100}+\frac{50}{100}\times 0.4\] \[=0.4+0.5\times 0.4\] \[=0.6\] Hence \[0.6=1-\frac{300}{500+x}\] \[\frac{300}{500+x}=0.4\] \[\Rightarrow \] \[500+x=\frac{300}{0.4}=750\] \[x=750-500\] \[=250K\]
You need to login to perform this action.
You will be redirected in
3 sec
