Kvant Physics Problem 15
The physical system is a refrigerator operating between the refrigerated compartment and the surrounding room.
Verified: yes
Verdicts: PASS + PASS
Solve time: 17m25s
Source on kvant.digital
Problem
Through the walls of a refrigerator, an amount of heat $Q=190\text{ ккал}$ enters per hour. The temperature inside the refrigerator is $T_1=+5^\circ\text{ C}$, while the temperature in the room is $T_2=+20^\circ\text{ C}$. What is the minimum power consumed by this refrigerator from the electrical mains?
V. N. Kopylov
Setup and Assumptions
The physical system is a refrigerator operating between the refrigerated compartment and the surrounding room. The unknown quantity is the minimum electrical power required to maintain the temperature difference.
The given data are
$$Q = 190\ \text{kcal/h},$$
where $Q$ is the amount of heat entering the refrigerator each hour through its walls,
$$T_1 = +5^\circ\text{C}, \qquad T_2 = +20^\circ\text{C},$$
where $T_1$ is the temperature inside the refrigerator and $T_2$ is the temperature of the room.
The refrigerator must continuously remove the incoming heat $Q$ from the cold compartment and reject it to the warmer surroundings.
The minimum possible power corresponds to an ideal reversible refrigerator operating on the Carnot cycle. Mechanical losses, electrical losses, heat leaks other than the stated one, and irreversibilities of all kinds are neglected. Absolute temperatures are used in all thermodynamic relations:
$$T_1 = 278\ \text{K}, \qquad T_2 = 293\ \text{K}.$$
Physical Principles
For a refrigerator, the coefficient of performance is defined as
$$\varepsilon = \frac{Q_1}{A},$$
where $Q_1$ is the heat extracted from the cold reservoir and $A$ is the work supplied to the refrigerator.
For a reversible Carnot refrigerator operating between temperatures $T_1$ and $T_2$,
$$\varepsilon_{\max} = \frac{T_1}{T_2-T_1}.$$
The minimum required work is obtained when the coefficient of performance is maximal. Hence
$$A_{\min} = \frac{Q_1}{\varepsilon_{\max}}.$$
Power is work per unit time:
$$P=\frac{A}{t}.$$
Since the heat entering through the walls each hour must be removed each hour in steady operation,
$$Q_1 = Q.$$
Derivation
For the ideal refrigerator,
$$\varepsilon_{\max} = \frac{T_1}{T_2-T_1}.$$
Substituting this expression into the definition of the minimum work gives
$$A_{\min} = Q,\frac{T_2-T_1}{T_1}.$$
The given temperatures in kelvin are
$$T_1 = 273+5 = 278\ \text{K},$$
$$T_2 = 273+20 = 293\ \text{K}.$$
Hence
$$A_{\min} = 190\ \text{kcal/h} \cdot \frac{293-278}{278}.$$
Since
$$293-278=15,$$
we obtain
$$A_{\min} = 190\cdot\frac{15}{278} \approx 10.25\ \text{kcal/h}.$$
This is the minimum work that must be supplied each hour.
To convert to power, first convert kilocalories per hour to watts. Using
$$1\ \text{kcal}=4186\ \text{J},$$
we find
$$A_{\min} = 10.25\times4186 \approx 4.29\times10^4\ \text{J/h}.$$
Therefore
$$P_{\min} = \frac{4.29\times10^4\ \text{J}}{3600\ \text{s}} \approx 11.9\ \text{W}.$$
Result
The minimum power consumption of a refrigerator operating between temperatures $T_1$ and $T_2$ is
$$P_{\min} = \frac{Q}{t}, \frac{T_2-T_1}{T_1}.$$
Substituting the given values,
$$P_{\min} = 190\ \text{kcal/h} \cdot \frac{15}{278} \cdot \frac{4186\ \text{J}}{1\ \text{kcal}} \cdot \frac{1}{3600\ \text{s/h}} \approx 11.9\ \text{W}.$$
Thus,
$$\boxed{P_{\min}\approx 12\ \text{W}}.$$
Sanity Checks
The factor
$$\frac{T_2-T_1}{T_1} = \frac{15}{278} \approx 0.054$$
is much smaller than unity. Consequently the required work is much smaller than the heat being pumped, which is characteristic of a refrigerator and agrees with the large coefficient of performance of a Carnot machine operating across only a $15\ \text{K}$ temperature difference.
Dimensional consistency is satisfied because the ratio $(T_2-T_1)/T_1$ is dimensionless. Multiplying a heat-flow rate by this dimensionless factor yields a work-flow rate, that is, a power.
If the temperature difference approached zero,
$$T_2-T_1 \to 0,$$
then
$$P_{\min}\to 0,$$
which matches the expectation that no work is needed to maintain equal temperatures. If the temperature difference increased while $T_1$ remained fixed, the required power would increase.
The most common source of error is using Celsius temperatures directly in the Carnot formula. The relation
$$\varepsilon_{\max} = \frac{T_1}{T_2-T_1}$$
requires absolute temperatures in kelvin; replacing $T_1=278\ \text{K}$ by $5^\circ\text{C}$ would produce a grossly incorrect result.