Question

An ideal heat engine is working between temperature ${T_1}$ and ${T_2}$ has efficiency $\eta$ . If both the temperatures are raised by ${100^ \circ }\,K$ each, the new efficiency will be
A. $\eta$
B. Less than $\eta$
C. More than $\eta$
D. Cannot be predicted.

Answer 1

In this problem, we need to find the efficiency of a heat engine in terms of already given efficiency. The efficiency of a heat engine is the ratio of the difference between the temperatures to the high temperature. Find the initial efficiency in terms of given values. For the second case find the efficiency when the temperature increases and thus compare both the efficiencies.

Complete step by step answer:
Let us consider that the temperature ${T_2}$ is greater than temperature ${T_1}$ , that is ${T_2} > {T_1}$ . The efficiency of an ideal heat engine is given as:
$\eta = \dfrac{{{T_2} - {T_1}}}{{{T_2}}}$
Here, $\eta$ has the same meaning as given in the question.
The temperatures ${T_2},{T_1}$ are in Kelvin.
As per the given condition, both the temperatures are raised by ${100^ \circ }\,K$ , the temperatures will be
${T_1}' = {T_1} + 10$
And ${T_2}' = {T_2} + 10$
Thus, the new efficiency $\eta '$ will be given as:
$\eta ' = \dfrac{{{T_2}' - {T_1}'}}{{{T_2}'}}$
The denominator is the higher temperature.
Substituting the values of new temperatures, we get
$\eta ' = \dfrac{{\left( {{T_2} + 100} \right) - \left( {{T_1} + 100} \right)}}{{{T_2} + 100}}$
$\Rightarrow \eta ' = \dfrac{{{T_2} - {T_1}}}{{{T_2} + 100}}$
Comparing it with the initial efficiency, we have
As the denominator value of efficiency $\eta '$ is greater than the initial efficiency, the magnitude of efficiency $\eta '$ will decrease as the temperatures rise. We know that as the denominator is increased, the magnitude decreases. Also, the denominator will be positive as the temperature is in the Kelvin scale which is an absolute scale.

So, the correct answer is “Option A”.

Note:
As the value of denominator increases, the overall value decreases and as the value of denominator decreases, the overall value increases.
Note that the temperature on the Kelvin and kelvin scale has non-negative values.
If we consider temperature ${T_2} < {T_1}$ , the same results will be obtained.