A Carnot engine operates between two reservoirs at 550 K and 250 K.It transfers 300 J to the reservoir at low temperature.The amount of work done is

Question

Quizplus · Accepted Answer

The efficiency of a Carnot engine is given by $ \eta = 1 - \frac{T_{C}}{T_{H}} $, where $ T_{C} $ is the temperature of the cold reservoir and $ T_{H} $ is the temperature of the hot reservoir. Here, $ T_{C} = 250 $ K and $ T_{H} = 550 $ K, so $ \eta = 1 - \frac{250}{550} = \frac{300}{550} $. The work done $ W $ is related to the heat rejected $ Q_{C} $ and the efficiency by $ W = Q_{H} - Q_{C} $, where $ Q_{H} $ is the heat absorbed from the hot reservoir. Since $ \eta = \frac{W}{Q_{H}} $, we can rearrange to find $ Q_{H} = \frac{W}{\eta} $. Given $ Q_{C} = 300 $ J, and knowing that $ Q_{H} = Q_{C} + W $, we can solve for $ W $ using the efficiency equation. Rearranging and solving gives $ W = \eta \cdot Q_{H} = \frac{300}{550} \cdot (Q_{C} + W) $. Solving this equation for $ W $ gives $ W = 360 $ J.

A Carnot Engine Operates Between Two Reservoirs at 550 K