Question

Carnot’s cycle is said to have $25\%$ efficiency when it operates between T(source) and
$300K$ (Sink). Temperature T is.
A) $200K$
B) $300K$
C) $400K$
D) None of these.

Answer 1

We know the theoretical maximum efficiency obtained when the heat engine operates between the two temperatures is called Carnot’s efficiency. Carnot’s efficiency can be calculated using the formula,
$\eta = \dfrac{{{T_2} - {T_1}}}{{{T_2}}}$
Where,
${T_1}$ is the temperature of the source.
The temperature of the sink is ${T_2}$.

Complete step by step answer:
We must remember that the Carnot’s theorem also referred to as Carnot’s rule was proposed by Nicolas Leonard Carnot within the year 1824, with the principle that there are limits on maximum efficiency for any given engine. It depends mostly on boiling and freezing reservoir temperatures.
Given,
The value of Carnot efficiency is $25\%$.
The temperature of the source is $300K$.
The Carnot’s efficiency is calculated by,
$\eta = \dfrac{{{\text{Network done per cycle}}}}{{{\text{Total amount of heat absorbed per cycle}}}}$
$\eta = \dfrac{{{Q_1} - {O_2}}}{{{Q_1}}}$
$\eta = 1 - \dfrac{{{Q_2}}}{{{Q_1}}}\xrightarrow{{}}1$
We know that heat is directly proportional to temperature thus,
$\dfrac{{{Q_2}}}{{{Q_1}}} = \dfrac{{{T_2}}}{{{T_1}}}$
Substituting the above equation 1
$\eta = 1 - \dfrac{{{T_2}}}{{{T_1}}}$
Now substitute known values in the above equation,
$\dfrac{{25}}{{100}} = 1 - \dfrac{{300K}}{{{T_1}}}$
On simplifying we get,
$\Rightarrow \dfrac{1}{4} = 1 - \dfrac{{300K}}{{{T_1}}}$
$\Rightarrow {T_1} = \dfrac{{300}}{{0.75}} = 400K$
Therefore, the option (C) is correct.

Note:
Now we discuss about the four steps involved in Carnot cycle as follows,
The first process is reversible isothermal gas expansion. In this process, the amount of heat absorbed by the ideal gas q from the heat source, which is at a temperature of T.
The second process is reversible adiabatic gas expansion. Now, the system is thermally protected, and the gas continues to expand and work is done on the surroundings. Now the temperature is lower, ${T_l}$ .
The third process is reversible isothermal gas compression process. Here, the heat loss ${q_{out}}$ happens only if the surroundings do the work at temperature ${T_l}$.
The fourth process is reversible adiabatic gas compression. Again the system is thermally insulated. The temperature again rises back to ${T_h}$ as the surroundings continue to do their work on the gas.