Kvant Physics Problem 188

Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m11s
Source on kvant.digital

Problem

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

Setup and Assumptions

The system consists of a refrigerator operating over a time interval $\tau$, consuming electrical power $W$, and a mass of water initially contained in a vessel at temperature $t^\circ\text{C}$ that is fully converted into ice.

The amount of water is $n$ liters. The density of water is taken as $1,\text{kg/L}$, so the mass is $m = n,\text{kg}$.

The final state of the substance is ice at $0^\circ\text{C}$. Heat exchange with the surroundings occurs only through the refrigerator condenser into the room. Heat losses other than those mediated by the refrigerator are neglected. The specific heat capacity of water is $c = 4.2 \times 10^3,\text{J/(kg·K)}$, and the latent heat of fusion of water is $L = 3.34 \times 10^5,\text{J/kg}$.

The required quantity is the total heat $Q_{\text{room}}$ released into the room during the process.

Physical Principles

The first law of thermodynamics for the refrigerator over the time interval $\tau$ gives the relation between heat delivered to the room $Q_{\text{room}}$, heat extracted from the cold body $Q_{\text{cold}}$, and the work input $A$:

$Q_{\text{room}} = Q_{\text{cold}} + A.$

The electrical work supplied to the refrigerator is

$A = W\tau.$

The heat removed from the water consists of two contributions. Cooling the water from $t$ to $0^\circ\text{C}$ requires

$Q_1 = mc\Delta T,$

and freezing the water at $0^\circ\text{C}$ requires

$Q_2 = mL.$

Thus the total heat extracted from the water is

$Q_{\text{cold}} = mc t + mL.$

Derivation

The mass of water is expressed through volume as $m = n,\text{kg}$.

The heat removed during cooling is computed from the temperature change $\Delta T = t,\text{K}$, since the Kelvin and Celsius increments are identical. This gives

$Q_1 = n c t.$

The heat removed during solidification is

$Q_2 = n L.$

The total heat extracted from the water becomes

$Q_{\text{cold}} = n(ct + L).$

The electrical energy supplied to the refrigerator over time $\tau$ is

$A = W\tau.$

Substituting these expressions into the energy balance for the refrigerator yields

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

Result

The heat released into the room is

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

Substituting numerical values,

$Q_{\text{room}} = n\left(4.2 \times 10^3,t + 3.34 \times 10^5\right),\text{J} + W\tau.$

With $n$ in $\text{kg}$ (or liters), $t$ in $\text{K}$ (or $^\circ\text{C}$), $W$ in watts, and $\tau$ in seconds, the result is

$\boxed{Q_{\text{room}} = n\left(4.2 \times 10^3 t + 3.34 \times 10^5\right) + W\tau ;\text{J}}.$

Sanity Checks

Dimensional consistency follows from each term being an energy: $nct$ has units $\text{kg}\cdot\text{J/(kg·K)}\cdot\text{K} = \text{J}$, $nL$ has units $\text{kg}\cdot\text{J/kg} = \text{J}$, and $W\tau$ has units $\text{W}\cdot\text{s} = \text{J}$.

In the limit $W \to 0$, the heat released into the room equals the heat removed from the water, corresponding to an idealized refrigerator with no electrical input. In the limit $t \to 0$, only latent heat contributes, reducing the expression to $Q_{\text{room}} = nL + W\tau$, consistent with freezing already-cooled water.

A sign error would most easily arise in the thermodynamic balance, where heat removed from the cold reservoir and work input both contribute positively to heat delivered to the room; reversing either term would violate energy conservation for the refrigerator cycle.