A carnot engine takes 3 × 10^6 cal of heat from a reservoir at 627^ , and gives it to a sink at 27^ . The work done by the engine is: - Physics MCQ AIEEE 2003 2026 | ExamTest.live

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A carnot engine takes 3 \times 1 0^{6} cal of heat from a reservoir at 62 7^{\circ} C, and gives it to a sink at 2 7^{\circ} C . The work done by the engine is:

Solution & Step-by-step Explanation

Temperatures in Kelvin: T_{1} = 627 + 273 = 900 K T_{2} = 27 + 273 = 300 K Efficiency η = 1 - \frac{T _{2}}{T _{1}} = 1 - \frac{300}{900} = \frac{2}{3} .Work done W = η Q_{1} : W = \frac{2}{3} \times (3 \times 1 0^{6} cal) = 2 \times 1 0^{6} cal Convert to Joules (using 1 cal = 4.2 J): W = 2 \times 1 0^{6} \times 4.2 = 8.4 \times 1 0^{6} J

Try it yourself before checking the explanation above.

A carnot engine takes 3 \times 1 0^{6} cal of heat from a reservoir at 62 7^{\circ} C, and gives it to a sink at 2 7^{\circ} C . The work done by the engine is:

4.2 \times 1 0^{6} J

8.4 \times 1 0^{6} J

16.8 \times 1 0^{6} J

Zero

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