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Refrigerator Or Heat Pump MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

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

  • 11 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 }

The temperature inside a refrigerator is $t_2^{\circ} \mathrm{C}$ and the room temperature is $t_1^{\circ} \mathrm{C}$. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be:

A Carnot engine, having the efficiency of a heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at a lower temperature is:

A refrigerator works between 4°C and 30°C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is: (Take 1 cal = 4.2 joules)

The coefficient of performance of a refrigerator is 5. If the temperature inside freezer is - 20°C, the temperature of the surroundings to which it rejects heat is:

A Carnot engine having an efficiency of $\frac{1}{10}$ as a heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at a lower temperature is:

A Carnot engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done (in J) in each cycle to operate the refrigerator is:

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A refrigerator freezes 1 kg of water at $0^{\circ} \mathrm{C}$ in 3 minutes. The room temperature is $27^{\circ} \mathrm{C}$. The latent heat of fusion of ice is 80 cal/g then choose the incorrect statement

In a refrigerator, one removes heat from a lower temperature and deposits it to the surroundings at a higher temperature. In this process, mechanical work has to be done, which is provided by an electrical motor. If the motor is of 1 kW power and heat is transferred from $-3^{\circ} \mathrm{C}$ to $27^{\circ} \mathrm{C}$, find the heat taken out of the refrigerator per second assuming its efficiency is $50 \%$ of a perfect engine.

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If the coefficient of performance of a refrigerator is 5 and 6 operates at the room temperature $\left(27^{\circ} \mathrm{C}\right)$, find the temperature inside the refrigerator.

Concepts Covered - 0

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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