question_answer

    A Carnot engine whose low temperature reservoir is at\[7{}^\circ C\] has an efficiency of 50%. It is desired to increase the efficiency to 70%. By how many degrees should the temperature of the high temperature reservoir be increased?

A)  840 K                                   

B)  280 K

C)  560 K                                   

D)  380 K

Correct Answer: D

Solution :

                Carnot engine is a reversible device which draws heat from a hotter reservoir at temperature\[{{T}_{1}}\], produces a positive amount of work in the surroundings and discharges rest amount of heat into a colder reservoir at temperature\[{{T}_{2}}\]. Efficiency is defined as                 \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\]                 \[\frac{50}{100}=1-\frac{273+7}{{{T}_{1}}}\] \[\Rightarrow \]               \[\frac{1}{2}=1-\frac{280}{{{T}_{1}}}\] \[\Rightarrow \]               \[\frac{280}{{{T}_{1}}}=\frac{1}{2}\] \[\Rightarrow \]               \[{{T}_{1}}=560\,K\] Let new temperature of high temperature reservoir is\[T_{1}^{}\]. Then,    \[\frac{70}{100}=1-\frac{280}{T_{1}^{}}\] \[\Rightarrow \]               \[\frac{280}{T_{1}^{}}=\frac{3}{10}\] \[\Rightarrow \]               \[T_{1}^{}=\frac{280\times 10}{3}=933\,K\] \[\therefore \]Increase in temperature \[=933-560=373K-380K\]