Kvant Physics Problem 188

Verified: yes
Verdicts: PASS + PASS
Solve time: 2m20s
Source on kvant.digital

Problem

A refrigerator with power $W$, over a time $\tau$, turned $n$ liters of water, which initially had a temperature $t^{\circ}~\text{C}$, into ice. How much heat was released into the room during this time?

Setup and Assumptions

A refrigerator operates for a time interval $\tau$ while consuming electrical power $W$. During this interval it freezes $n$ liters of water initially at temperature $t^\circ\text{C}$. The task is to determine the total heat released into the room during this process.

The water is treated as incompressible with density $1,\text{kg}/\text{L}$, so its mass is $m = n,\text{kg}$. The initial temperature $t$ satisfies $t \ge 0^\circ\text{C}$, and the final state of the water is ice at $0^\circ\text{C}$. The specific heat capacity of water is denoted $c$, and the latent heat of fusion of ice is denoted $\lambda$. Heat losses other than those associated with the refrigerator and the water are neglected. All energy ultimately rejected by the refrigerator condenser is assumed to enter the room.

Physical Principles

The heat required to cool liquid water from $t$ to $0^\circ\text{C}$ is given by

$Q_1 = mc t.$

The heat released during phase transition from liquid water at $0^\circ\text{C}$ to ice at $0^\circ\text{C}$ is

$Q_2 = m\lambda.$

The first law of thermodynamics for the refrigerator operating in a cycle states that the heat delivered to the hot reservoir (the room) equals the sum of the heat extracted from the cold reservoir and the mechanical work supplied:

$Q_{\text{room}} = Q_{\text{removed}} + A,$

where $A$ is the electrical work input over time $\tau$, given by

$A = W\tau.$

Derivation

The total heat removed from the water consists of two contributions. The first is cooling the water from $t^\circ\text{C}$ to $0^\circ\text{C}$, which yields $Q_1 = mc t$. The second is freezing the water at $0^\circ\text{C}$, which yields $Q_2 = m\lambda$. The total extracted heat is therefore

$Q_{\text{removed}} = mc t + m\lambda.$

Substituting this into the energy balance for the refrigerator gives the heat released into the room:

$Q_{\text{room}} = mc t + m\lambda + W\tau.$

Replacing the mass by $m = n$ (with $n$ in liters interpreted as kilograms) gives

$Q_{\text{room}} = n c t + n\lambda + W\tau.$

Result

The total heat released into the room is

$Q_{\text{room}} = n(ct + \lambda) + W\tau.$

Substituting units explicitly,

Q_{\text{room}} = n,(\text{kg}) \cdot \left(c,\frac{\text{J}}{\text{kg}\cdot^\circ\text{C}} \cdot t,^\circ\text{C} + \lambda,\frac{\text{J}}{\text{kg}}\right) + W,\text{W} \cdot \tau,\text{s}.

Thus,

$\boxed{Q_{\text{room}} = n(ct + \lambda) + W\tau ;\text{J}}.$

Sanity Checks

Dimensional consistency follows from each term carrying units of energy: $ct$ has units $\text{J}/\text{kg}$, $\lambda$ has units $\text{J}/\text{kg}$, and multiplication by $n$ in kilograms produces joules; the term $W\tau$ also has units $\text{J}$ since $\text{W} = \text{J}/\text{s}$.

In the limiting case $W \to 0$, the result reduces to the purely thermodynamic heat release from cooling and freezing the water, $n(ct + \lambda)$. If $t \to 0$, only latent heat remains, giving $n\lambda + W\tau$. For large operating time $\tau$, the contribution $W\tau$ dominates, consistent with the refrigerator continuously dumping both extracted heat and input work into the room.