Question

Efficiency of a Carnot engine is 50% and the temperature of the sink is 500K. If the temperature of source is kept constant and its efficiency raised to 60% then the required temperatures of the sink will be
(A) $100K$
(B) $600K$
(C) $400K$
(D) $500K$

Answer 1

Hint: Here we have to find the required temperature of the sink if efficiency of Carnot engine is \[50\% \] and the constant temperature of the source means sink is\[500{\text{ }}K\]. If efficiency of Carnot engine is raised to \[60\% \].
Efficiency is basically used for measuring how much work or energy we can conserve in a process. In other words, it is like comparing the output of the energy to the input of the energy in any given system.

Complete step by step solution:
When the efficiency of the Carnot engine is $50\% $, then the outlet temperature ${T_2}$ is $500K$ that is ${T_2} = 500K$
We know the efficiency formula that is
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
Here we the value of efficiency that is $50\% $ and the value of ${T_2}$ is $500K$ now substitute the values in above equation
$ \Rightarrow \dfrac{{50}}{{100}} = 1 - \dfrac{{500}}{{{T_1}}}$
Now we have to find the value of ${T_1}$ hence it to R.H.S so the equation becomes
$ \Rightarrow \dfrac{{500}}{{{T_1}}} = 1 - \dfrac{{50}}{{100}}$
Now take the L.C.M so we get
$ \Rightarrow \dfrac{{500}}{{{T_1}}} = \dfrac{{100 - 50}}{{100}}$
So after the further simplification we get
${T_1} = \dfrac{{500 \times 100}}{{50}}$

After calculating we get
${T_1} = 1000K$
Now after increase in the efficiency up to $60\% $ with the temperature ${T_1} = 1000K$
Then we know the efficiency formula
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
Now substitute the values in the above equation
$\dfrac{{60}}{{100}} = 1 - \dfrac{{{T_2}}}{{1000}}$
Here we have to find ${T_2}$so take it outside then the equation becomes
$\dfrac{{{T_2}}}{{1000}} = 1 - \dfrac{{60}}{{100}}$
Now take the LCM in R.H.S
$\dfrac{{{T_2}}}{{1000}} = \dfrac{{100 - 60}}{{100}}$
Now after further simplification we get
${T_2} = \dfrac{{1000 \times 40}}{{100}}$
After calculating we get
${T_2} = 400K$
It is the required temperature of sink.

Hence, the correct answer is option (C).

Note: A Carnot heat engine is a theoretical engine that operates on the Carnot cycle it is an ideal reversible heat engine especially as postulated in the statement of Carnot's principle of engine efficiency