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In a refrigerator, heat is transferred from a lowertemperature medium (the refrigerated space) to a highertemperature one (the kitchen air). Is this a violation of the second law of thermodynamics? Explain.

Short Answer

Expert verified
Answer: No, the heat transfer process in a refrigerator from a low-temperature medium to a high-temperature one does not violate the second law of thermodynamics. This is because the refrigerator consumes external work to accomplish the heat transfer, which results in an increase in the total entropy of the system.

Step by step solution

01

Recall the Second Law of Thermodynamics

The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible. In terms of heat transfer, it implies that heat cannot spontaneously flow from a colder body to a warmer body without any external work.
02

Understand the refrigerator's working principle

A refrigerator works on the principle of a heat pump, which uses external work to transfer heat from a low-temperature medium (refrigerated space) to a high-temperature medium (kitchen air). The refrigerator carries out this process through a cyclic process, consisting of compression, condensation, expansion, and evaporation of a refrigerant.
03

Determine if the refrigerator operation violates the second law

The refrigerator transfers heat from a lower-temperature medium to a higher-temperature one, but it does so by using external work. As a result, the total entropy of the system (refrigerator, refrigerated space, and the kitchen) increases. Therefore, the refrigerator's operation does not violate the second law of thermodynamics, as the overall entropy of the system increases during the process.
04

Conclusion

The process of transferring heat from a low-temperature medium to a high-temperature medium in a refrigerator does not violate the second law of thermodynamics. This is because the refrigerator consumes external work to accomplish the heat transfer, which results in an increase in the total entropy of the system.

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Most popular questions from this chapter

The \(\mathrm{COP}\) of a refrigerator decreases as the temperature of the refrigerated space is decreased. That is, removing heat from a medium at a very low temperature will require a large work input. Determine the minimum work input required to remove \(1 \mathrm{kJ}\) of heat from liquid helium at \(3 \mathrm{K}\) when the outside temperature is 300 K.

An air-conditioning system is used to maintain a house at a constant temperature of \(20^{\circ} \mathrm{C}\). The house is gaining heat from outdoors at a rate of \(20,000 \mathrm{kJ} / \mathrm{h},\) and the heat generated in the house from the people, lights, and appliances amounts to \(8000 \mathrm{kJ} / \mathrm{h}\). For a COP of \(2.5,\) determine the required power input to this air-conditioning system.

A heat pump supplies heat energy to a house at the rate of \(140,000 \mathrm{kJ} / \mathrm{h}\) when the house is maintained at \(25^{\circ} \mathrm{C} .\) Over a period of one month, the heat pump operates for 100 hours to transfer energy from a heat source outside the house to inside the house. Consider a heat pump receiving heat from two different outside energy sources. In one application the heat pump receives heat from the outside air at \(0^{\circ} \mathrm{C} .\) In a second application the heat pump receives heat from a lake having a water temperature of \(10^{\circ} \mathrm{C}\). If electricity costs \(\$ 0.105 / \mathrm{kWh}\), determine the maximum money saved by using the lake water rather than the outside air as the outside energy source.

Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at \(1400 \mathrm{K}\) and rejects the waste heat to another reservoir at temperature \(T\) The second engine receives this energy rejected by the first one, converts some of it to work, and rejects the rest to a reservoir at \(300 \mathrm{K}\). If the thermal efficiencies of both engines are the same, determine the temperature \(T .\)

Consider a Carnot refrigerator and a Carnot heat pump operating between the same two thermal energy reservoirs. If the COP of the refrigerator is \(3.4,\) the COP of the heat pump is \((a) 1.7\) (b) 2.4 \((c) 3.4\) \((d) 4.4\) \((e) 5.0\)

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