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    How to Score High in JEE Main 2026 Physics – Tips, Tricks & Important Topics

    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.

    • 8 Questions around this concept.

    Solve by difficulty

    Which is correct about function of refrigerator

    The coefficient of performance of a Carnot refrigerator working between $30^{\circ} \mathrm{Cand} 0^{\circ} \mathrm{C}$ is

    Concepts Covered - 1

    Refrigerator or Heat Pump

    A refrigerator or heat pump is basically a heat engine running 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 $\left(\beta\right)$ -

      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.
      i.e. $W=0$ so that $Q_1=Q_2$ and hence $\beta=\infty$
      - Carnot refrigerator-

      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 $\mathrm{T}_1=$ temperature of surrounding, $\mathrm{T}_2=$ temperature of cold body and $T_1>T_2$
      when $\mathrm{T}_2=0$ 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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