Question

The efficiency of a Carnot's heat engine is \( \frac{1}{3} \). If the temperature of the source is decreased by \(50^\circ \text{C}\) and the temperature of the sink is increased by \(25^\circ \text{C}\), the efficiency of the engine becomes \( \frac{3}{16} \). The initial temperature of the sink is

• A. 325 K
• B. 375 K
• C. 350 K
• D. 300 K

Hint:

Use the Carnot efficiency formula \( \eta = 1 - \frac{T_C}{T_S} \) and relate given changes to form two equations. Eliminate \(T_S\) using substitution to solve for \(T_C\).

Answer 1

The Correct Option is D

Step 1: Let the initial temperature of the source be \( T_S \) and of the sink be \( T_C \). From the Carnot efficiency formula: \[ \eta = 1 - \frac{T_C}{T_S} \] Given: \[ \frac{1}{3} = 1 - \frac{T_C}{T_S} \Rightarrow \frac{T_C}{T_S} = \frac{2}{3} \Rightarrow T_S = \frac{3}{2} T_C
\text{(1)} \] Now, after changes: \[ T_S' = T_S - 50,
T_C' = T_C + 25 \] New efficiency: \[ \eta' = 1 - \frac{T_C + 25}{T_S - 50} = \frac{3}{16} \Rightarrow \frac{T_C + 25}{T_S - 50} = \frac{13}{16}
\text{(2)} \] Substitute (1) in (2): \[ \frac{T_C + 25}{\frac{3}{2}T_C - 50} = \frac{13}{16} \] Step 2: Solve the equation: \[ 16(T_C + 25) = 13\left(\frac{3}{2}T_C - 50\right) \Rightarrow 16T_C + 400 = \frac{39}{2}T_C - 650 \Rightarrow 800 + 1300 = 39T_C - 32T_C \Rightarrow 7T_C = 2100 \Rightarrow T_C = 300\,\text{K} \] % Final Answer \[ \boxed{300\,\text{K}} \]