AP EAMCET · PHYSICS · Thermodynamics
The efficiency of a Carnot's heat engine is \(\frac{1}{3}\). If the temperature of the source is decreased by \(50^{\circ} \mathrm{C}\) and the temperature of the sink is increased by \(25^{\circ} \mathrm{C}\), 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
Answer & Solution
Correct Answer
(D) 300 K
Step-by-step Solution
Detailed explanation
\(\eta_1 = 1 - \frac{T_L}{T_S} \Rightarrow \frac{1}{3} = 1 - \frac{T_L}{T_S}\) \(\frac{T_L}{T_S} = \frac{2}{3} \Rightarrow T_S = \frac{3}{2}T_L\) \(\eta_2 = 1 - \frac{T_L+25}{T_S-50} \Rightarrow \frac{3}{16} = 1 - \frac{T_L+25}{T_S-50}\) \(\frac{T_L+25}{T_S-50} = \frac{13}{16}\)…
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