An ideal Carnot engine operates between a high temperature reservoir at $492 \mathrm {~K}$ and a river with water at $290 \mathrm {~K} .$ If it absorbs $4000 \mathrm {~J}$ of energy each cycle,how much work per cycle does it perform?

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.In this case, $T_c = 290 \mathrm {~K}$ and $T_h = 492 \mathrm {~K}$, so:$$\eta = 1 - \frac{290 \mathrm {~K}}{492 \mathrm {~K}} = 0.411$$The work done per cycle is given by:$$W = Q_h \eta$$where $Q_h$ is the heat absorbed by the engine per cycle.In this case, $Q_h = 4000 \mathrm {~J}$, so:$$W = 4000 \mathrm {~J} \times 0.411 = 1644 \mathrm {~J}$$Therefore, the correct answer is A.

An Ideal Carnot Engine Operates Between a High Temperature Reservoir 492 K492 \mathrm {~K}492 K

An Ideal Carnot Engine Operates Between a High Temperature Reservoir $492 \mathrm {~K}$