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Refrigerator Or Heat Pump - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Refrigerator or Heat Pump is considered one the most difficult concept.

  • 2 Questions around this concept.

Solve by difficulty

For a refrigerator which of the following statements is true 

1) A refrigerator is basically a heat engine running in the reverse direction 

2) Coefficient of performance \beta = Heat extracted / Work done 

3) The relation b/w coefficient of performance and efficiency of the refrigerator is 

\eta = \frac{1 - \beta }{\beta }

Concepts Covered - 1

Refrigerator or Heat Pump

A refrigerator or heat pump is basically a heat engine run in the reverse direction.

It consists of three parts
1. Source: At higher temperature T1
2. Working substance: It is called refrigerant. I.e liquid ammonia and freon works as a working substance.
3. Sink: At lower temperature T2.

  • Working of refrigerator

     

As shown in the above figure, The working substance takes heat Q2 from a sink (contents of refrigerator) at lower temperature T2, has a net amount of work done W on it by an external agent (usually compressor of refrigerator) and gives out a larger amount of heat Q1 to a hot body at temperature T1 (usually atmosphere).

  • Use of refrigerator-

  The cold body is cooled more and more with the help of a refrigerator. Because the refrigerator transfers heat from a cold to a hot body at the expense of mechanical energy supplied to it by an external agent.

  • Coefficient of performance (\beta)-

The coefficient of performance is defined as the ratio of the heat extracted from the cold body to the work needed to transfer it to the hot body.

\beta=\frac{\text { Heat extracted }}{\text { work done }}=\frac{Q_{2}}{W}=\frac{Q_{2}}{Q_{1}-Q_{2}}

A perfect refrigerator is one which transfers heat from cold to a hot body without doing work.

\text { i.e. } W=0 \text { so that } Q_{1}=Q_{2} \text { and hence } \beta=\infty

  • Carnot refrigerator-

         \text { For Carnot refrigerator } \frac{Q_{1}}{Q_{2}}=\frac{T_{1}}{T_{2}}

   \therefore \frac{Q_{1}-Q_{2}}{Q_{2}}=\frac{T_{1}-T_{2}}{T_{2}} \text { or } \frac{Q_{2}}{Q_{1}-Q_{2}}=\frac{T_{2}}{T_{1}-T_{2}}

So using \beta= \frac{Q_{2}}{Q_{1}-Q_{2}}

we get \beta=\frac{T_{2}}{T_{1}-T_{2}}

where T1 = temperature of surrounding, T2 = temperature of cold body and  T_{1}> T_{2}

  when T2 = 0 then \beta=0

I.e  if the cold body is at the temperature equal to absolute zero, then the coefficient of performance will be zero

  • The relation between  \beta  and \eta of the refrigerator

            \beta=\frac{1-\eta}{\eta}

 

 

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Refrigerator or Heat Pump

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Reference Books

Refrigerator or Heat Pump

Physics Part II Textbook for Class XI

Page No. : 313

Line : 65

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