Question

A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52 K, its efficiency is doubled. The temperature in Kelvin of the source will be ________ .

Hint:

When dealing with problems involving changes in Carnot efficiency, set up two separate equations for the initial and final states. This creates a system of two equations with two unknowns (T₁ and T₂), which can then be solved simultaneously.

Answer 1

Correct Answer: 208

The efficiency ($\eta$) of a heat engine is the ratio of work done (W) to the heat input ($Q_H$).
Given that it converts one-fourth of the heat input into work, the initial efficiency is $\eta_1 = \frac{W}{Q_H} = \frac{1}{4}$.
For a reversible engine (Carnot engine), the efficiency is also given by $\eta = 1 - \frac{T_{sink}}{T_{source}} = 1 - \frac{T_2}{T_1}$.
Case 1: $\eta_1 = \frac{1}{4}$.
$\frac{1}{4} = 1 - \frac{T_2}{T_1} \implies \frac{T_2}{T_1} = 1 - \frac{1}{4} = \frac{3}{4}$. (Equation 1)
Case 2: The sink temperature is reduced by 52 K, so the new sink temperature is $T_2' = T_2 - 52$. The efficiency doubles, so $\eta_2 = 2 \times \eta_1 = 2 \times \frac{1}{4} = \frac{1}{2}$.
$\frac{1}{2} = 1 - \frac{T_2'}{T_1} = 1 - \frac{T_2 - 52}{T_1}$.
$\frac{T_2 - 52}{T_1} = 1 - \frac{1}{2} = \frac{1}{2}$.
$T_2 - 52 = \frac{T_1}{2}$. (Equation 2)
From Equation 1, we have $T_2 = \frac{3}{4}T_1$. Substitute this into Equation 2:
$\frac{3}{4}T_1 - 52 = \frac{1}{2}T_1$.
$\frac{3}{4}T_1 - \frac{2}{4}T_1 = 52$.
$\frac{1}{4}T_1 = 52$.
$T_1 = 52 \times 4 = 208$ K.
The temperature of the source is 208 K.