Question

A Carnot engine works between ice point and steam point. In order to increase the efficiency of the engine by $20\;\%$ , its source temperature will have to be changed by
(A) $29.5\;K$ increase
(B) $29.5\;K$ decrease
(C) ${37.5^\circ }C$ increase
(D) ${29.5^\circ }C$ decrease

Answer 1

We will use the given conditions to obtain the efficiency of the engine. Then we shall increase the obtained efficiency by $20\;\%$ . This efficiency is obtained when the source temperature has been raised which can be obtained by the same formula used to find the efficiency.

Formula used
$\eta = 1 - \dfrac{{{T_1}}}{{{T_2}}}$
where ${T_1}$ is the temperature of the sink and ${T_2}$ is the temperature of the source.

Complete Step by step solution
The efficiency of a Carnot cycle is given by
$\eta = 1 - \dfrac{{{T_1}}}{{{T_2}}}$
where ${T_1}$ is the lower temperature and ${T_2}$ is the higher temperature.
Here both the temperatures must be in Kelvin. Thus ${T_1}$ here is the temperature at which ice melts which is $273\;K$ and ${T_2}$ is the temperature at which water starts boiling off into steam which is $373\;K$ .
Thus substituting the values of the temperatures in the efficiency formula, we get,
$\Rightarrow \eta = 1 - \dfrac{{273\;K}}{{373\;K}}$
Upon further solving we get,
$\Rightarrow \eta = 1 - 0.7319$
$\Rightarrow \eta = 0.2681$
It is given that the efficiency is to be increased by $20\;\%$ . Therefore the new efficiency after the increase is $120\;\%$ of the original efficiency of the Carnot engine.
$\Rightarrow {\eta _{new}} = \dfrac{{120}}{{100}} \times 0.2681$
$\Rightarrow {\eta _{new}} = 0.3217$
This is the new efficiency of the engine. In the question, it is said that the source temperature needs to be changed. Therefore the sink temperature should remain the same. Thus the sink temperature ${T_1}$ remains $273\;K$ . However, the higher temperature of the source which was previously ${T_2}$ needs to be changed to ${T_3}$ .
Applying the formula of efficiency of the Carnot engine, we get
$\Rightarrow {\eta _{new}} = 1 - \dfrac{{{T_1}}}{{{T_3}}}$
Substituting the respective values, we get,
$\Rightarrow 0.3217 = 1 - \dfrac{{273\;K}}{{{T_3}}}$
Upon rearranging the equation we get,
$\Rightarrow \dfrac{{273\;K}}{{{T_3}}} = 1 - 0.3217$
$\Rightarrow \dfrac{{273\;K}}{{{T_3}}} = 0.6783$
On rearranging the terms, we get,
$\Rightarrow {T_3} = \dfrac{{273\;K}}{{0.6783}}$
$\Rightarrow {T_3} = 402.48\;K$
Therefore we can see that the source temperature has increased.
We need to find the amount by which it has increased. Thus we will subtract the initial source temperature from the newly obtained final source temperature, i.e. ${T_3} - {T_2}$ .
$\Rightarrow {T_3} - {T_2} = 402.48\;K - 373\;K$
$\Rightarrow {T_3} - {T_2} = 29.48\;K$
Thus the correct answer is option (A).

Note
Here in the options provided, we have both temperatures in Kelvin and Celsius scales. It should be known that both these scales have the same difference in one unit change in their temperatures. This means that ${1^\circ }C$ rise or fall in temperature would mean the same when the temperature is increased or decreased by $1\;K$ .