Q76 / 90 · JEE Main 2023 (08 Apr Shift 2) · Physics

Work done by a Carnot engine operating between temperatures $127 ° C$ and $27 ° C$ is $2 kJ$ . The amount of heat transferred to the engine by the reservoir is:

Correct answer
A
$8 kJ$
B
$2.67 kJ$
C
$2 kJ$
D
$4 kJ$

Solution

The temperatures in Kelvin scales are
$T_{2} = 27 ° C + 273 = 300 K T_{1} = 127 ° C + 273 = 400 K$
The efficiency of the Carnot engine is
$η = 1 - \frac{T_{2}}{T_{1}} = 1 - \frac{300}{400} = \frac{1}{4}$
Using the relation, $η = \frac{W}{Q}$
$\Rightarrow \frac{2 kJ}{Q} = \frac{1}{4} \Rightarrow Q = 8 kJ$

Hence, the correct answer is option: 'A'.

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Work done by a Carnot engine operating between temperatures 127°C and 27°C is 2 kJ. The amount of heat transferred to the engine by the reservoir is:

Solution

Relevant Questions

Work done by a Carnot engine operating between temperatures $127 ° C$ and $27 ° C$ is $2 kJ$ . The amount of heat transferred to the engine by the reservoir is: