# 9.5 Applications of thermodynamics: heat pumps and refrigerators  (Page 5/7)

 Page 5 / 7

## Section summary

• An artifact of the second law of thermodynamics is the ability to heat an interior space using a heat pump. Heat pumps compress cold ambient air and, in so doing, heat it to room temperature without violation of conservation principles.
• To calculate the heat pump’s coefficient of performance, use the equation ${\text{COP}}_{\text{hp}}=\frac{{Q}_{\text{h}}}{W}$ .
• A refrigerator is a heat pump; it takes warm ambient air and expands it to chill it.

## Conceptual questions

Explain why heat pumps do not work as well in very cold climates as they do in milder ones. Is the same true of refrigerators?

In some Northern European nations, homes are being built without heating systems of any type. They are very well insulated and are kept warm by the body heat of the residents. However, when the residents are not at home, it is still warm in these houses. What is a possible explanation?

Why do refrigerators, air conditioners, and heat pumps operate most cost-effectively for cycles with a small difference between ${T}_{\text{h}}$ and ${T}_{\text{c}}$ ? (Note that the temperatures of the cycle employed are crucial to its $\text{COP}$ .)

Grocery store managers contend that there is less total energy consumption in the summer if the store is kept at a low temperature. Make arguments to support or refute this claim, taking into account that there are numerous refrigerators and freezers in the store.

Can you cool a kitchen by leaving the refrigerator door open?

## Problem exercises

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

4.82

Suppose you have an ideal refrigerator that cools an environment at $-\text{20}\text{.}0\text{º}\text{C}$ and has heat transfer to another environment at $\text{50}\text{.}0\text{º}\text{C}$ . What is its coefficient of performance?

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

0.311

In a very mild winter climate, a heat pump has heat transfer from an environment at $5\text{.}\text{00}\text{º}\text{C}$ to one at $\text{35}\text{.}0\text{º}\text{C}$ . What is the best possible coefficient of performance for these temperatures? Explicitly show how you follow the steps in the Problem-Solving Strategies for Thermodynamics .

(a) What is the best coefficient of performance for a heat pump that has a hot reservoir temperature of $\text{50}\text{.}0\text{º}\text{C}$ and a cold reservoir temperature of $-\text{20}\text{.0ºC}$ ? (b) How much heat transfer occurs into the warm environment if $3\text{.60}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{J}$ of work ( $\text{10}\text{.}0\text{kW}\cdot \text{h}$ ) is put into it? (c) If the cost of this work input is $\text{10.0 cents/kW}\cdot \text{h}$ , how does its cost compare with the direct heat transfer achieved by burning natural gas at a cost of 85.0 cents per therm. (A therm is a common unit of energy for natural gas and equals $1\text{.}\text{055}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{J}$ .)

(a) 4.61

(b) $1\text{.}\text{66}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{J}\phantom{\rule{0.25em}{0ex}}\text{or 3}\text{.}\text{97}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{kcal}$

(c) To transfer $1\text{.}\text{66}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{J}$ , heat pump costs $1.00, natural gas costs$1.34.

(a) What is the best coefficient of performance for a refrigerator that cools an environment at $-\text{30}\text{.}0\text{º}\text{C}$ and has heat transfer to another environment at $\text{45}\text{.}0º\text{C}$ ? (b) How much work in joules must be done for a heat transfer of 4186 kJ from the cold environment? (c) What is the cost of doing this if the work costs 10.0 cents per $3\text{.}\text{60}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{J}$ (a kilowatt-hour)? (d) How many kJ of heat transfer occurs into the warm environment? (e) Discuss what type of refrigerator might operate between these temperatures.

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

$\text{27.6ºC}$

An ideal heat pump is being considered for use in heating an environment with a temperature of $\text{22}\text{.}0\text{º}\text{C}$ . What is the cold reservoir temperature if the pump is to have a coefficient of performance of 12.0?

A 4-ton air conditioner removes $5\text{.}\text{06}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{J}$ (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 $3\text{.}\text{60}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{J}$ (one kilowatt-hour)? (c) Discuss whether this cost seems realistic. Note that the energy efficiency rating ( $\text{EER}$ ) of an air conditioner or refrigerator is defined to be the number of British thermal units of heat transfer from a cold environment per hour divided by the watts of power input.

(a) $1\text{.}\text{44}×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{J}$

(b) 40 cents

(c) This cost seems quite realistic; it says that running an air conditioner all day would cost \$9.59 (if it ran continuously).

Show that the coefficients of performance of refrigerators and heat pumps are related by ${\text{COP}}_{\text{ref}}={\text{COP}}_{\text{hp}}-1$ .

Start with the definitions of the $\text{COP}$ s and the conservation of energy relationship between ${Q}_{\text{h}}$ , ${Q}_{\text{c}}$ , and $W$ .

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