The fridge in the room

Problem

A domestic fridge is placed in the middle of a perfectly isolated room and switched on.

• (a) In the first few hours, does the room temperature $T$ fall, rise, or stay constant? Justify.
• (b) After a few hours the door is opened and left wide open. How does $T(t)$ evolve? Hint: apply energy conservation to a control volume enclosing the whole room.
• (c) Estimate the temperature increase after $4$ hours, assuming a room of $50\;\text{m}^3$ and an average dissipated power of $100\;\text{W}$.

Solution

(a) T rises. The room is thermally isolated, but energy continuously enters through the mains socket. A fridge does not “destroy” heat: it pumps it from inside the compartment to the back (the condensers). Since everything stays inside the isolated room, the electrical energy absorbed ends up entirely as heat in the room’s air. So $T$ rises.

(b) T keeps rising, more steeply. With the door open the motor works harder: it tries to pump heat from one environment into the same environment, an impossible task that keeps it running continuously. More electrical work absorbed → more heat dissipated → the curve $T(t)$ becomes steeper, never decreasing.

(c) Estimate. Apply energy conservation to the “room” control volume. Energy input in $4$ hours: $E = P\,t = 100 \cdot 4 \cdot 3600 = 1{,}44\cdot 10^6\;\text{J}$ Mass of air ($\rho \approx 1{,}2\;\text{kg/m}^3$): $m = \rho V \approx 1{,}2 \cdot 50 = 60\;\text{kg}$ With the specific heat of air at constant volume $c_v \approx 718\;\text{J/(kg·K)}$: $\Delta T = \frac{E}{m\,c_v} = \frac{1{,}44\cdot 10^6}{60 \cdot 718} \approx 33\;\text{K}$ This is an “air only” estimate: accounting for walls and furniture that absorb heat, the real increase would be smaller.

$\ev{T\text{ rises in both cases; steeper with the door open; }\Delta T \sim 30\;\text{K}}$

Reworked from (Povey 2015, §12.3).

Links

Topics: Thermodynamics Concepts: First law of thermodynamics Skills: Limiting-case and thought-experiment analysis Objects: Fridge