If the co-efficient of performance of a refrigerator is 5 and operates at the room temperature (27 °C), find t

NCERT Exemplar · Q21

Q. If the co-efficient of performance of a refrigerator is 5 and operates at the room temperature (27 °C), find the temperature inside the refrigerator.

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Concept understanding — Refrigerator Coefficient of Performance

A refrigerator does not "create cold." It moves heat from a cold place (the inside of the fridge) to a hot place (the kitchen). That is the first thing to hold in your mind. The cold side gets colder because heat is being pulled out of it; the hot side gets hotter because that same heat is dumped there, plus the energy used to do the work.

If you put a hot bowl of soup inside the fridge, the fridge has to work harder. Why? Because more heat needs to be moved out. The "performance" of a fridge is simply: how much heat did I manage to pull out of the cold space, for every unit of work (electricity) I put in?

That is the intuition. You want a big "heat removed" for a small "work paid."

The precise statement

Let:

$Q_c$ = heat extracted from the cold reservoir (the inside of the fridge). This is the "useful" effect.
$Q_h$ = heat dumped into the hot reservoir (the kitchen). This is always larger than $Q_c$ .
$W$ = work input (electrical energy consumed by the compressor).

From the first law of thermodynamics (energy conservation), for a complete cycle:

Q_h = Q_c + W

The work you put in ends up as extra heat added to the hot side. So the Coefficient of Performance (COP) of a refrigerator is defined as:

\text{COP}_R = \frac{\text{What you want}}{\text{What you pay}} = \frac{Q_c}{W}

Using $W = Q_h - Q_c$ , we get the standard form:

\boxed{\text{COP}_R = \frac{Q_c}{Q_h - Q_c}}

Important

COP is not efficiency. Efficiency (for a heat engine) is always less than 1. COP for a refrigerator is always greater than 1 (often 2–6 for real fridges). Why? Because $Q_c$ can be several times larger than $W$ . You are moving heat, not converting it into work.

A concrete example

Suppose a fridge extracts 200 J of heat from the inside ( $Q_c = 200$ J) and dumps 250 J into the kitchen ( $Q_h = 250$ J). Then the work done is $W = 250 - 200 = 50$ J.

\text{COP} = \frac{200}{50} = 4

This means: for every 1 J of electrical energy you pay for, the fridge moves 4 J of heat out of your food. That is a COP of 4 — quite good.

If the fridge were perfect (impossible), it would move heat with zero work, and COP would be infinite. Real fridges have COP values between 2 and 6, depending on the temperature difference they have to work against.

Watch out

A common mistake: thinking COP = $Q_h/W$ or $Q_h/(Q_h - Q_c)$ . That is the COP of a heat pump (used for heating), not a refrigerator. For a fridge, the numerator is always $Q_c$ , the heat removed from the cold space.

Why the formula makes physical sense …

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The coefficient of performance of a refrigerator relates the heat extracted from the cold reservoir to the work done. For an ideal refrigerator, this can be expressed in terms of the absolute temperatures of the hot and cold reservoirs. Using the given coefficient of performance and room temperature, the temperature inside the refrigerator is found to be $\boxed{-23 \text{ °C}}$ .

A refrigerator's purpose is to transfer heat from a colder region (inside the refrigerator) to a warmer region (the room). This process requires external work, as heat naturally flows from hot to cold, not the other way around. The efficiency of this process is quantified by the Coefficient of Performance (COP).

The Coefficient of Performance ( $K$ ) for a refrigerator is defined as the ratio of the heat extracted from the cold reservoir ( $Q_C$ ) to the work input ( $W$ ):

K = \frac{Q_C}{W}

For an ideal refrigerator (one operating on a reversible Carnot cycle), the COP can also be expressed in terms of the absolute temperatures of the cold reservoir ( $T_C$ ) and the hot reservoir ( $T_H$ ):

K = \frac{T_C}{T_H - T_C}

This formula is derived from the principles of thermodynamics, specifically the Carnot cycle, and represents the maximum possible COP for any refrigerator operating between these two temperatures. In the absence of information suggesting a non-ideal refrigerator, we assume ideal operation for such problems.

Important

All temperatures used in thermodynamic formulas, especially those involving ratios or differences like the COP, must be expressed in absolute temperature units (Kelvin).

Let's apply this understanding to solve the problem.

Identify the given values and convert units:
- The coefficient of performance of the refrigerator, $K = 5$ .
- The room temperature (hot reservoir temperature), $T_H = 27 \text{ °C}$ .
- We must convert $T_H$ to Kelvin: $T_H = 27 + 273 = 300 \text{ K}$
State the relevant formula:

The formula relating the coefficient of performance to the absolute temperatures of the hot and cold reservoirs for an ideal refrigerator is:
$K = \frac{T_C}{T_H - T_C}$

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Q.If the co-efficient of performance of a refrigerator is 5 and operates at the room temperature (27 °C), find the temperature inside the refrigerator.

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