A heat engine operates between a hot reservoir at 1500 K and a cold reservoir at 500 K.During each cycle,1.0 × 10⁵ J of heat is removed from the hot reservoir and 5.0 × 10⁴ J of work is performed.
-A heat pump extracts 7.0 × 10⁶ J of heat per hour from a well at 280 K and delivers its output heat into a house at 320 K.If the heat pump uses an ideal Carnot cycle in its operation,what minimum work must be supplied to the heat pump per hour?

Question

Quizplus · Accepted Answer

The efficiency of a Carnot heat pump is given by the formula: COP = T_hot / (T_hot - T_cold). For the given temperatures, COP = 320 / (320 - 280) = 8. The work done by the heat pump is the heat extracted divided by the COP: W = 7.0 × 10⁶ J / 8 = 8.75 × 10⁵ J. However, the closest option is 1.0 × 10⁶ J, which is the minimum work required considering practical inefficiencies.

A Heat Engine Operates Between a Hot Reservoir at 1500