Question

A Carnot engine has efficiency $\dfrac{1}{5}$ . Efficiency becomes $\dfrac{1}{3}$ when temperature of the sink is decreased by $50K$. What is the temperature of sink?
A. $325\,K$
B. $375\,K$
C. $300\,K$
D. $350\,K$

Answer 1

Hint: Here we will use the formula of the efficiency of the Carnot engine to calculate the temperature of the sink. Here, we will calculate the temperature ${T_1}$ in terms of ${T_2}$ , and then we will calculate the value of ${T_2}$ by substitution method. After calculating the value of ${T_2}$, we will put it in the equation of ${T_1}$ to calculate the temperature of the sink ${T_1}$ .

Complete step by step answer:
We know that the efficiency of the Carnot engine is defined as the ratio of the work done to obtain the output from the engine to the heat supplied to the engine and is given by
$\eta = \dfrac{W}{{{Q_1}}}$
$ \Rightarrow\eta = \dfrac{{{Q_1} - {Q_2}}}{{{Q_1}}}$
$ \Rightarrow\eta = 1 - \dfrac{{{Q_2}}}{{{Q_1}}}$
Also, we can show that
$\dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{{T_2}}}{{{T_1}}}$
Therefore, the efficiency of the Carnot engine will become
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
Now, the efficiency of the Carnot engine is $\dfrac{1}{5}$ as given in the question.
Therefore, $\dfrac{1}{5} = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
$ \Rightarrow \,\dfrac{1}{5} = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}$
$ \Rightarrow \,{T_1} = 5{T_1} - 5{T_2}$
$ \Rightarrow \,4{T_1} = 5{T_2}$
$ \Rightarrow \,{T_1} = \dfrac{{5{T_2}}}{4}$
Now, it is given in the question that when the temperature is reduced to $50\,K$, the efficiency of the Carnot engine will become $\dfrac{1}{3}$, hence, we will take the temperature ${T_2}$ as ${T_2} - 50$ .
Therefore, $\dfrac{1}{3} = 1 - \dfrac{{{T_2} - 50}}{{{T_1}}}$
 $ \Rightarrow \,\dfrac{1}{3} = \dfrac{{{T_1} - {T_2} - 50}}{{{T_1}}}$
 $ \Rightarrow \,{T_1} = 3{T_1} - 3{T_2} - 150$
$ \Rightarrow \,2{T_1} - 3{T_2} - 150 = 0$
Now, putting the values of ${T_1}$ in the above equation, we get
$2\left( {\dfrac{{5{T_2}}}{4}} \right) - 3{T_2} - 150 = 0$
$ \Rightarrow \,5{T_2} - 6{T_2} - 300 = 0$
$ \Rightarrow \, - {T_2} = 300$
Taking magnitude, we get
${T_2} = 300\,K$
Now, we will calculate the value of ${T_1}$ by putting the value of ${T_2}$ as shown below
${T_1} = \dfrac{{5 \times 300}}{4}$
$ \Rightarrow \,{T_1} = 375\,K$
Which is the required temperature.
Hence, the temperature of the sink will be $375\,K$.

So, the correct answer is “Option B”.

Note:
The efficiency of the Carnot engine can never be $100\% $ because if the efficiency will be $100\% $ then $\eta = 1$ , therefore, we will get ${Q_2} = 0$ which means that the heat from the source can be converted to work done. Therefore the temperature of the sink will be greater than unity which is a violation of the second law of thermodynamics. Hence, the efficiency of the Carnot engine can never be $100\% $.