Question:
A Carnot engine is working between a source at constant temperature \(T_1\) and a sink at temperature \(T_2\). Its efficiency is \(\frac{1}{8}\). Upon decreasing the temperature of the sink by 50°C, the efficiency becomes \(\frac{1}{4}\). Find \(T_1\) and \(T_2\).
A Carnot engine is working between a source at constant temperature \(T_1\) and a sink at temperature \(T_2\). Its efficiency is \(\frac{1}{8}\). Upon decreasing the temperature of the sink by 50°C, the efficiency becomes \(\frac{1}{4}\). Find \(T_1\) and \(T_2\).
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Always convert °C to K in thermodynamics before applying Carnot efficiency.
Updated On: Jun 20, 2026
- \(T_1 = 127^\circ C, \; T_2 = 77^\circ C\)
- \(T_1 = 400^\circ C, \; T_2 = 315^\circ C\)
- \(T_1 = 215^\circ C, \; T_2 = 100^\circ C\)
- \(T_1 = 100^\circ C, \; T_2 = 215^\circ C\)
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The Correct Option is A
Solution and Explanation
Step 1: Carnot efficiency formula.
\[ \eta = 1 - \frac{T_2}{T_1} \] Temperatures must be in Kelvin.
Step 2: First condition.
\[ \frac{1}{8} = 1 - \frac{T_2}{T_1} \] \[ \frac{T_2}{T_1} = \frac{7}{8} \] \[ T_2 = \frac{7}{8}T_1 \]
Step 3: Second condition (sink reduced by 50°C).
\[ T_2' = T_2 - 50 \] New efficiency: \[ \frac{1}{4} = 1 - \frac{T_2 - 50}{T_1} \] \[ \frac{T_2 - 50}{T_1} = \frac{3}{4} \] \[ T_2 - 50 = \frac{3}{4}T_1 \]
Step 4: Solve equations.
Substitute \(T_2 = \frac{7}{8}T_1\): \[ \frac{7}{8}T_1 - 50 = \frac{3}{4}T_1 \]
Step 5: Simplify.
\[ \frac{7}{8}T_1 - \frac{3}{4}T_1 = 50 \] \[ \left(\frac{7}{8} - \frac{6}{8}\right)T_1 = 50 \] \[ \frac{1}{8}T_1 = 50 \] \[ T_1 = 400 K \]
Step 6: Find \(T_2\).
\[ T_2 = \frac{7}{8} \times 400 = 350 K \] Convert to Celsius: \[ T_1 = 127^\circ C, \quad T_2 = 77^\circ C \] \[ \boxed{127^\circ C, 77^\circ C} \]
\[ \eta = 1 - \frac{T_2}{T_1} \] Temperatures must be in Kelvin.
Step 2: First condition.
\[ \frac{1}{8} = 1 - \frac{T_2}{T_1} \] \[ \frac{T_2}{T_1} = \frac{7}{8} \] \[ T_2 = \frac{7}{8}T_1 \]
Step 3: Second condition (sink reduced by 50°C).
\[ T_2' = T_2 - 50 \] New efficiency: \[ \frac{1}{4} = 1 - \frac{T_2 - 50}{T_1} \] \[ \frac{T_2 - 50}{T_1} = \frac{3}{4} \] \[ T_2 - 50 = \frac{3}{4}T_1 \]
Step 4: Solve equations.
Substitute \(T_2 = \frac{7}{8}T_1\): \[ \frac{7}{8}T_1 - 50 = \frac{3}{4}T_1 \]
Step 5: Simplify.
\[ \frac{7}{8}T_1 - \frac{3}{4}T_1 = 50 \] \[ \left(\frac{7}{8} - \frac{6}{8}\right)T_1 = 50 \] \[ \frac{1}{8}T_1 = 50 \] \[ T_1 = 400 K \]
Step 6: Find \(T_2\).
\[ T_2 = \frac{7}{8} \times 400 = 350 K \] Convert to Celsius: \[ T_1 = 127^\circ C, \quad T_2 = 77^\circ C \] \[ \boxed{127^\circ C, 77^\circ C} \]
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