15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators

What is the coefficient of performance of an ideal heat pump that has heat transfer from a cold temperature of $- \text{25} . 0 º \text{C}$ to a hot temperature of $\text{40} . 0 º \text{C}$?

What is the best coefficient of performance possible for a hypothetical refrigerator that could make liquid nitrogen at $- \text{200} º \text{C}$ and has heat transfer to the environment at $\text{35} . 0 º \text{C}$?

(a) What is the best coefficient of performance for a heat pump that has a hot reservoir temperature of $\text{50} . 0 º \text{C}$ and a cold reservoir temperature of $- \text{20} .\text{0}º\text{C}$? (b) How much heat transfer occurs into the warm environment if $3 .\text{60} \times \text{10}^{7} \text{ J}$ of work ($\text{10} . 0 \text{ kW} \cdot \text{h}$) is put into it? (c) If the cost of this work input is $\text{10}.\text{0 cents}/\text{kW} \cdot \text{h}$, how does its cost compare with the direct heat transfer achieved by burning natural gas at a cost of 85.0 cents per therm…

Suppose you want to operate an ideal refrigerator with a cold temperature of $- \text{10} . 0º \text{C}$, and you would like it to have a coefficient of performance of 7.00. What is the hot reservoir temperature for such a refrigerator?

A 4-ton air conditioner removes $5 . \text{06} \times \text{10}^{7} \text{ J}$ (48,000 British thermal units) from a cold environment in 1.00 h. (a) What energy input in joules is necessary to do this if the air conditioner has an energy efficiency rating ($\text{EER}$) of 12.0? (b) What is the cost of doing this if the work costs 10.0 cents per $3 . \text{60} \times \text{10}^{6} \text{ J}$ (one kilowatt-hour)? (c) Discuss whether this cost seems realistic. Note that the energy efficiency rating…