Refrigerator Or Heat Pump

A refrigerator or heat pump operates on the principle of moving heat from a cooler to a warmer area, effectively controlling the temperature of a space. This concept is essential for understanding thermodynamics and is highly relevant to students preparing for board exams and competitive exams like JEE and NEET. By exploring how these devices work, you can apply thermodynamic laws to solve practical problems and understand everyday appliances better. This article explains the functioning of refrigerators and heat pumps in simple terms and includes a solved example to show these principles in action, making complex ideas accessible and applicable.

What is a Refrigerator or Heat Pump?

A refrigerator and a heat pump are both devices that use similar principles of thermodynamics but serve different purposes.

Refrigerator: A refrigerator is a household appliance designed to keep food and beverages cold by removing heat from its interior and expelling it to the surrounding environment. It operates based on the refrigeration cycle, where a refrigerant absorbs heat inside the fridge and releases it outside through coils. This process maintains a low temperature inside the refrigerator, ensuring that perishable items remain fresh for longer periods.

Heat Pump: A heat pump is a versatile climate control system used for both heating and cooling. It works by transferring heat from one place to another using a refrigerant cycle similar to that of a refrigerator. In heating mode, a heat pump extracts heat from the outside air (even in cold conditions) and transfers it indoors to warm a space. In cooling mode, it reverses the process, removing heat from the indoor air and releasing it outside. Heat pumps are known for their efficiency in managing indoor temperatures year-round.

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

Working of Refrigerator

A refrigerator works by employing the principles of thermodynamics to transfer heat from its interior to the external environment, thereby keeping its contents cool. Here’s a step-by-step explanation of its working process:

1. Evaporation:

   • Inside the refrigerator, there is an evaporator coil containing a refrigerant. This refrigerant is a liquid at a low temperature and pressure.
   • As the refrigerant flows through the evaporator coil, it absorbs heat from the interior of the refrigerator. This absorption causes the refrigerant to evaporate and turn into a gas.

2. Compression:

   • The gaseous refrigerant then flows to the compressor, which is typically located at the back or bottom of the refrigerator.
   • The compressor compresses the gas, increasing its pressure and temperature.

3. Condensation:

   • The high-pressure, high-temperature gas then moves through the condenser coils, usually located on the back or bottom of the refrigerator.
   • As the gas passes through the condenser coils, it releases heat to the surrounding environment and condenses back into a liquid state.

4. Expansion:

   • The high-pressure liquid refrigerant then passes through an expansion valve or capillary tube. This component reduces the pressure of the refrigerant.
   • The pressure drop causes the refrigerant to expand and cool significantly, making it ready to absorb heat again.

5. Cycle Repeats:

   • The cooled refrigerant returns to the evaporator coil, and the cycle begins anew.

Throughout this process, the refrigerator’s interior remains cooler because the heat is continuously being transferred away from it to the external environment. This efficient cycle ensures that the contents of the refrigerator stay fresh and preserved.

As shown in the above figure, The working substance takes heat Q₂ from a sink (contents of the refrigerator) at lower temperature T₂, has a net amount of work done W on it by an external agent (usually compressor of the 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 a cold to a hot body without doing work.

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

Carnot Refrigerator

A Carnot refrigerator is a theoretical model that represents the most efficient possible refrigeration system, based on the principles of thermodynamics. Named after the French physicist Sadi Carnot, this idealized refrigerator operates on the Carnot cycle, which is a thermodynamic cycle that provides the highest efficiency for heat transfer.

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 $T_2=0$ then $\beta=0$

i.e if the cold body is at a 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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Solved Examples Based on Refrigerator or Heat Pump

Example 1: 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

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

1) 1,2 and 3

2) 1 and 2

3) 2 and 3

4) only 2

Solution:

Refrigerator

It is a device that is used to keep bodies at a temperature lower than their surroundings.

Relation between

$\beta$ and $\eta$ is $\beta=\frac{1-\eta}{\eta}$

Hence, the answer is the option (2).

Example 2: A Carnot engine, having the efficiency of $\eta=1 / 10$ as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed (in joules) from the reservoir at a lower temperature is

1) 90

2) 99

3) 100

4) 1

Solution:

Coefficient of performance (β)

$
\beta=\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2}
$
wherein
$
T_1>T_2
$

For a perfect refrigerator $\beta \rightarrow \infty$
$
\eta=\frac{1}{10}
$

Coefficient of performance of refrigerator,
$
\begin{aligned}
& \beta=\frac{1}{\eta}-1=9 \\
& \beta=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2} \\
& 9=\frac{Q_2}{10 J}
\end{aligned}
$

Hence, the answer is the option (1).

Example 3: A Carnot’s 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 :

1) 420

2) 772

3) 2100

4) 2520

Solution:

Coefficient of performance ( $\beta$ )
$
\beta=\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2}
$
wherein
$
T_1>T_2
$

For a perfect refrigerator $\beta \rightarrow \infty$

$\begin{aligned} & \text { Coefficient of performance }=\beta=\frac{Q_2}{W}=\frac{Q_2}{Q_1-Q_2}=\frac{T_2}{T_1-T_2} \\ & \frac{250}{50}=\frac{500}{W} \\ & \Rightarrow W=100 \mathrm{cal} \\ & \text { 1cal=4.2 J } \\ & \mathrm{W}=420 \mathrm{~J} \\ & \end{aligned}$.

Hence, the answer is the option (1).

Example 4: If the minimum possible work is done by a refrigerator in converting 100 grams of water at $0^{\circ} \mathrm{C}$ to ice, how much heat ( in calories) is released to the surroundings at temperature (Latent heat of ice ) to the nearest integer?

1) 8791

2) 309

3) 4567

4) 21

Solution

$\begin{aligned} & \mathrm{w}+\mathrm{Q}_1=\mathrm{Q}_2 \\ & \mathrm{w}=\mathrm{Q}_2-\mathrm{Q}_1 \\ & \text { C.O.P. }=\frac{\mathrm{Q}_1}{\mathrm{w}}=\frac{\mathrm{Q}_1}{\mathrm{Q}_2-\mathrm{Q}_1}=\frac{273}{300-273}=\frac{\mathrm{Q}_1}{\mathrm{w}} \\ & \mathrm{w}=\frac{27}{273} \times 80 \times 100 \times 4.2 \\ & \mathrm{Q}_2=\mathrm{w}+\mathrm{Q}_1 \\ & \mathrm{Q}_2=\left(\frac{27}{273} \times 80 \times 100 \times 4.2\right)+(80 \times 100 \times 4.2) \\ & \mathrm{Q}_2=\frac{300}{273} \times 80 \times 100=8791.2 \mathrm{cal} \approx 8791 \mathrm{cal}\end{aligned}$

Hence, the answer is the option (1).

Summary

Refrigerators and heat pumps, though based on similar thermodynamic principles, serve distinct functions: refrigerators cool items by transferring heat from their interior to the environment, while heat pumps provide versatile climate control by transferring heat for both heating and cooling. The Carnot refrigerator represents an ideal model, operating with maximum efficiency defined by the Carnot cycle. Its performance, quantified by the coefficient of performance (COP), highlights the theoretical limits of refrigeration efficiency. Practical examples illustrate how COP and thermodynamic efficiency influence real-world applications and energy consumption.