Daily Physicist

Takaaki Kajita image courtesy of Science Council of Japan under the Creative Commons Attribution 4.0 International License.

Today is the birthday of

• Takaaki Kajita (1959-).

Today’s Problem

At our local Sam’s Club it costs $2.48 for 16 pounds (7.27 kg) of ice. How many kilowatt-hours (kWh) is required to take 16 pounds of water at 20\(^{\circ}\)C (68\(^{\circ}\)F) down to ice at -20\(^{\circ}\)C (-4\(^{\circ}\)F), assuming 100% efficiency?

Is $2.48 a good deal considering the average price of a kWh of electricity is 17 cents?

Answer

The specific heat capacity of liquid water is \(c_p(H_2O\,\text{liq})=4.1844\,\text{kJ/kg}\cdot^\circ\text{K}\).
The specific heat capacity of solid water is \(c_p(H_2O\,\text{sol})=2.0\,\text{kJ/kg}\cdot^\circ\text{K}\).
The latent heat of fusion for water is \(L(H_2O\,\text{fus})=333.55\,\text{kJ/kg}\).
The total energy required is

\[E(H_2O)|_{+20^{\circ}C}^{-20^{\circ}C}=m\left[L(H_2O\,\text{fus})+\Delta T(\text{liq})c_p(H_2O\,\text{liq})+\Delta T(\text{sol})c_p(H_2O\,\text{sol})\right]\]

\[\Delta T(\text{liq})=\Delta T(\text{sol})=20^\circ\text{K}\]

Plugging in the numbers we get

\[E(H_2O)|_{+20^{\circ}C}^{-20^{\circ}C}=3324\,\text{kJ}\]

from the problem of March 6, 2026 we know that

\[1\,\text{kW}\cdot\text{hr}=1000\,\text{W}\cdot\text{hr}=1000\,\text{J/s}\cdot\text{hr}\left(\frac{3600\,\text{s}}{\text{hr}}\right)=3.6\times 10^6\,\text{J}\]

So 3324 kJ = 0.923 kWh or about 1 kWh, so it’s not a good deal, but we assumed 100% efficiency, and we’re not considering the convenience of ready to go ice.