If a refrigerator's door is kept open, will the room become cool or hot? Explain.

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Q. If a refrigerator's door is kept open, will the room become cool or hot? Explain.

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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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A refrigerator acts as a heat pump, consuming electrical energy to move heat. When its door is open, it expels more heat into the room than it removes from it, causing the room to become hotter.

To understand why a room becomes hotter when a refrigerator door is kept open, we need to consider how a refrigerator works and the fundamental principles of thermodynamics, specifically the First Law of Thermodynamics (conservation of energy).

A refrigerator is essentially a heat pump. Its purpose is to transfer heat from a colder region (the inside of the refrigerator) to a warmer region (the surrounding room). This process does not happen spontaneously; it requires an input of energy, typically electrical energy, to do work.

Here's a step-by-step explanation:

Refrigerator's Basic Function: A refrigerator works by absorbing heat from its interior (the cold compartment) and releasing that heat into its surroundings (the room). This is achieved through a refrigerant cycle involving compression, condensation, expansion, and evaporation.
Energy Input (Work Done): To move heat from a colder place to a warmer place, the refrigerator's compressor does work. This work is supplied by the electrical energy it consumes. Let's denote the heat absorbed from the cold interior as $Q_C$ and the work done by the compressor as $W$ .
Heat Expelled to the Room: According to the First Law of Thermodynamics, energy must be conserved. The total heat expelled to the warmer surroundings, $Q_H$ , is the sum of the heat absorbed from the cold interior and the work done by the compressor:

Q_H = Q_C + W

This means that the refrigerator always expels *more* heat into the room than it absorbs from its interior. The extra heat comes directly from the electrical energy consumed by the compressor, which is converted into heat.

4. Effect When the Door is Open:

* When the refrigerator door is open, the refrigerator tries to cool the entire room (or at least a significant portion of it) to its set internal temperature. …

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Q.If a refrigerator's door is kept open, will the room become cool or hot? Explain.

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