Question

An ideal heat engine operates in Carnot cycle between 127\(^\circ\)C and 27\(^\circ\)C. It absorbs \( 5 \times 10^4 \) cal of heat at higher temperature. Amount of heat converted to work is

• A. \( 4.8 \times 10^4 \) cal
• B. \( 2.4 \times 10^4 \) cal
• C. \( 1.25 \times 10^4 \) cal
• D. \( 6 \times 10^4 \) cal

Hint:

In Carnot engines, use the efficiency formula \( \eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \) to calculate the work done by the engine.

Answer 1

The Correct Option is C

The efficiency of a Carnot engine is given by: \[ \eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}} \] where \( T_{\text{hot}} = 127(^\circ)C = 127 + 273 = 400 \, \text{K} \) and \( T_{\text{cold}} = 27(^\circ)C = 27 + 273 = 300 \, \text{K} \). \[ \eta = 1 - \frac{300}{400} = 0.25 \] The work done by the engine is: \[ W = \eta Q_{\text{in}} = 0.25 \times 5 \times 10^4 = 1.25 \times 10^4 \, \text{cal} \] Thus, the amount of heat converted to work is \( 1.25 \times 10^4 \, \text{cal} \).