AP EAMCET · PHYSICS · Thermodynamics
A Carnot engine having efficiency \(60 \%\) receives heat from a source at a temperature 600 K. For the same sink temperature, to increase its efficiency to \(80 \%\), the temperature of the source is
- A 300 K
- B 900 K
- C 1200 K
- D 720 K
Answer & Solution
Correct Answer
(C) 1200 K
Step-by-step Solution
Detailed explanation
\( \eta_1 = 1 - \frac{T_L}{T_{H1}} \) \( 0.60 = 1 - \frac{T_L}{600 \text{ K}} \) \( T_L = (1 - 0.60) \times 600 \text{ K} = 240 \text{ K} \) \( \eta_2 = 1 - \frac{T_L}{T_{H2}} \) \( 0.80 = 1 - \frac{240 \text{ K}}{T_{H2}} \)…
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