JEE Physics 2023 Question with Solution

Given: A Carnot engine has initial efficiency $\frac{1}{3}$. When the cold reservoir temperature is increased by $x$, the efficiency becomes $\frac{1}{6}$. The hot reservoir temperature is $99^\circ \text{C}$.

Find: The value of $x$.

For a Carnot engine,

$$\eta = 1 - \frac{T_C}{T_H}$$

First convert the hot reservoir temperature to Kelvin:

$$T_H = 99 + 273 = 372 \, \text{K}$$

Initially,

$$\frac{1}{3} = 1 - \frac{T_C}{372}$$

So,

$$\frac{T_C}{372} = 1 - \frac{1}{3} = \frac{2}{3}$$

Hence,

$$T_C = \frac{2}{3} \times 372 = 248 \, \text{K}$$

After raising the cold reservoir temperature by $x$, the new efficiency is $\frac{1}{6}$:

$$\frac{1}{6} = 1 - \frac{T_C + x}{T_H}$$

Thus,

$$\frac{T_C + x}{372} = 1 - \frac{1}{6} = \frac{5}{6}$$

So,

$$T_C + x = \frac{5}{6} \times 372 = 310 \, \text{K}$$

Substituting $T_C = 248 \, \text{K}$,

$$248 + x = 310$$

Therefore,

$$x = 310 - 248 = 62 \, \text{K}$$

So the value of $x$ is $62 \, \text{K}$. The solution working gives Option A, although the listed option containing $62 \, \text{K}$ is D.

Given: $\eta_1 = \frac{1}{3}$, $\eta_2 = \frac{1}{6}$, and $T_H = 372 \, \text{K}$.

Find: Increase in cold reservoir temperature, $x$.

Using the Carnot relation,

$$1 - \eta = \frac{T_C}{T_H}$$

Initially,

$$1 - \frac{1}{3} = \frac{T_C}{372} = \frac{2}{3}$$

Hence,

$$T_C = 372 \times \frac{2}{3} = 248 \, \text{K}$$

After the increase,

$$1 - \frac{1}{6} = \frac{T_C + x}{372} = \frac{5}{6}$$

So,

$$T_C + x = 372 \times \frac{5}{6} = 310 \, \text{K}$$

Therefore,

$$x = 310 - 248 = 62 \, \text{K}$$

Thus, the required increase is $62 \, \text{K}$.