Q.A domestic refrigerator keeps its interior cold by continuously removing heat from inside and rejecting it, along with some extra energy, to the kitchen. Explain qualitatively, using the second law of thermodynamics, why this transfer of heat from a colder space to a warmer one requires external work to be supplied, and cannot happen by itself.

With 25% efficiency, the refrigerator's coefficient of performance is 3.

For a refrigerator, the coefficient of performance is:

$\beta = \dfrac{Q_2}{W} = \dfrac{Q_2}{Q_1 - Q_2}$

where $Q_2$ is heat extracted from the cold reservoir and $W = Q_1 - Q_2$ is the work input. The efficiency of the equivalent heat engine is $\eta = \dfrac{W}{Q_1} = 1 - \dfrac{Q_2}{Q_1}$.

…

Using COP = Q_cold / W, the work needed to remove 300 J of heat with a refrigerator of COP 4 is W = 300/4 = 75 J.

For a refrigerator, the Coefficient of Performance (COP) is defined as the ratio of heat removed from the cold reservoir (interior) to the work input required to do so:

COP = Q_cold / W

Given: COP = 4, Q_cold = 300 J.

Rearranging for W:

W = Q_cold / COP = 300 J / 4 = 75 J

…

Leaving the refrigerator door open while it runs heats the room rather than cooling it, because the compressor's work input adds more heat to the room (via the condenser coils) than is extracted from the room air by the evaporator coils.

A refrigerator works as a heat engine in reverse (a heat pump): it uses external work W (electrical energy driving the compressor) to extract heat Q2 from a cold reservoir (the inside of the fridge) and reject heat Q1 = Q2 + W into a hot reservoir (the room, via the condenser coils/grille at the back).

With the door open, the 'inside' of the fridge is no longer thermally separated from the room -- the evaporator keeps absorbing heat Q2 from the room air (since there is no closed insulated compartment anymore), and the compressor keeps rejecting Q1 = Q2 + W back into the same room through the condenser coils.

…

Leaving the refrigerator door open makes the room hotter rather than cooler, because the fridge dumps more heat into the room (through its condenser coils) than it removes from the air passing through it, the extra amount being the electrical work it consumes.

A refrigerator is a heat engine run in reverse (a heat pump). It absorbs heat Q2 from inside its cabinet at a lower temperature and, using electrical work input W, rejects a larger amount of heat Q1 = Q2 + W into the room through the condenser coils at the back, which is always warmer than the room.

With the door closed, this Q2 is drawn only from the small enclosed cabinet air, while Q1 is dumped into the room — the fridge cools its own interior while slightly warming the room.

…

COP = $\dfrac{T_2}{T_1-T_2}$ depends on the temperatures of the reservoirs, so it changes whenever those temperatures change — it is not a fixed constant of the refrigerator.

The coefficient of performance (COP) of a refrigerator (working as a reversed Carnot engine) is:

$\beta = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}$

where $T_2$ is the absolute temperature of the cold reservoir (inside the refrigerator) and $T_1$ is the absolute temperature of the hot reservoir (the room/surroundings).

…

Using COP = Qc/W with Qc = 200 J and COP = 3, the work required is W = 200/3 = 66.67 J.

For a refrigerator, the coefficient of performance (COP) is defined as the ratio of the heat Qc extracted from the cold reservoir (the refrigerator's interior) to the work W that must be supplied to run the refrigerator:

COP = Qc / W

Given Qc = 200 J and COP = 3:

W = Qc / COP = 200 / 3 = 66.666... J is approximately 66.67 J

…

The statement is True: an open refrigerator door does not cool the room — it actually heats it, because the compressor's motor keeps doing work and the total heat rejected to the room (heat removed + compressor's work) exceeds what would otherwise happen with the door shut.

A refrigerator is a heat engine run in reverse (a heat pump): it uses external work (done by the compressor motor) to extract heat Q_cold from inside (the cold reservoir) and reject a larger amount of heat Q_hot = Q_cold + W to outside (the room, the hot reservoir), as required by the second law of thermodynamics and conservation of energy.

Normally (door closed), the heat rejected to the room goes through the coils at the back, while the food compartment inside stays cool, isolated from the room.

If the door is left open:

• The refrigerator keeps trying to cool the room air that enters through the open door.
• It extracts heat from the room air (Q_cold) and, using the compressor's work (W), rejects Q_hot = Q_cold + W back into the same room through its rear coils. …