Kvant Physics Problem 15
A refrigerator maintains its internal air at temperature $T_1 = 5^\circ\text{C}$ while it is placed in a room at temperature $T_2 = 20^\circ\text{C}$.
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Solve time: 1m00s
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Problem
Heat $Q=190\text{ ккал}$ penetrates through the walls of a refrigerator in one hour. The temperature inside the refrigerator is $T_1=+5^\circ\text{ C}$, and in the room it is $T_2=+20^\circ\text{ C}$. What is the minimum power the refrigerator consumes from the mains?
V. N. Kopylov
Setup and Assumptions
A refrigerator maintains its internal air at temperature $T_1 = 5^\circ\text{C}$ while it is placed in a room at temperature $T_2 = 20^\circ\text{C}$. Heat penetrates through the walls into the interior at a rate corresponding to $Q = 190,\text{kcal per hour}$. The refrigerator must remove this heat continuously in steady state.
The unknown quantity is the minimum electrical power $P_{\min}$ required from the mains to sustain this steady operation.
The refrigerator is modeled as a reversible (Carnot) heat pump operating between two thermal reservoirs at fixed temperatures $T_1$ and $T_2$. All heat ingress is assumed to be exactly compensated by refrigeration work. Mechanical, electrical, and thermal losses beyond the ideal thermodynamic limit are neglected.
Physical Principles
The first law of thermodynamics for a refrigerator operating in steady state implies equality between heat removed from the cold reservoir per unit time and the heat influx into the system.
For a reversible refrigerator operating between temperatures $T_1$ and $T_2$, the coefficient of performance is given by
$$\mathrm{COP} = \frac{T_1}{T_2 - T_1},$$
where temperatures are measured in kelvin.
The coefficient of performance relates the cooling power $\dot{Q}$ and the mechanical power input $P$ through
$$\mathrm{COP} = \frac{\dot{Q}}{P}.$$
Derivation
The heat leakage into the refrigerator per unit time equals the cooling power required to maintain the internal temperature,
$$\dot{Q} = 190,\text{kcal/h}.$$
Conversion to SI units uses $1,\text{kcal} = 4184,\text{J}$, giving
$$\dot{Q} = 190 \times 4184,\text{J/h} = 7.9496 \times 10^5,\text{J/h}.$$
Conversion from hours to seconds yields
$$\dot{Q} = \frac{7.9496 \times 10^5}{3600},\text{W} = 2.208 \times 10^2,\text{W}.$$
The absolute temperatures are
$$T_1 = 278,\text{K}, \quad T_2 = 293,\text{K}.$$
The reversible coefficient of performance becomes
$$\mathrm{COP} = \frac{278}{293 - 278} = \frac{278}{15}.$$
The minimum required power follows from $P = \dot{Q}/\mathrm{COP}$,
$$P_{\min} = \dot{Q}\frac{T_2 - T_1}{T_1}.$$
Substitution gives
$$P_{\min} = 220.8,\text{W} \cdot \frac{15}{278}.$$
Result
The minimum electrical power is
$$P_{\min} = \frac{220.8 \cdot 15}{278},\text{W} = 11.9,\text{W}.$$
$$\boxed{P_{\min} \approx 12,\text{W}}$$
Sanity Checks
Dimensional consistency follows from $\dot{Q}$ expressed in watts and the coefficient of performance being dimensionless, so the resulting unit is watt.
The temperature difference $T_2 - T_1 = 15,\text{K}$ is small compared with the absolute temperature scale, producing a large coefficient of performance $\mathrm{COP} \approx 18.5$, consistent with a low theoretical work requirement.
The magnitude of $12,\text{W}$ is significantly smaller than the thermal leakage power of approximately $221,\text{W}$, which aligns with the expectation that an ideal refrigerator moves heat much more efficiently than direct conversion of heat into work.