How this calculation works
Cooling water to zero degrees does not by itself turn it all into ice. Freezing removes additional latent heat while the ideal phase-change temperature stays fixed. Cooling the resulting ice below zero adds another sensible-heat stage.
Removed kJ = mass × [4.186 × initial °C + 333.55 + 2.1 × (−final °C)]; duration = heat/useful cooling rateA worked example
One kilogram starting at twenty degrees needs 83.72 kJ removed to reach zero. Freezing removes 333.55 kJ, and cooling the ice to minus ten removes another twenty-one. Total removal is 438.27 kJ, or about 0.121742 thermal kWh. At 100 useful thermal watts the ideal duration is about 1.21742 hours; COP two gives about 0.060871 electrical kWh under the entered assumptions.
Good to know
Pure-water constant-property ideal model only; no container, heat ingress, nucleation, pressure effect or equipment selection. Real freezing duration and appliance input require actual operating data. This page does not recommend freezer sizing or food freezing procedures.
Reference: OpenStax — phase change and latent heat (opens in a new tab) ↗
What this calculation answers
Cooling water to zero degrees does not by itself turn it all into ice. Freezing removes additional latent heat while the ideal phase-change temperature stays fixed. Cooling the resulting ice below zero adds another sensible-heat stage.
This page budgets those three stages for pure water with constant material properties and an assumed zero-degree melting boundary. It is a thermal-energy calculation, not a direct prediction of a household appliance cycle.
The Water and ice equilibrium calculator asks a different question: how a finite warm-water and cold-ice mixture shares conserved energy without an external cooler.
Use the right input conventions
Cooling capacity means useful heat removal in thermal watts. It is not the electrical rating printed on a compressor or freezer. Enter COP separately when comparing ideal electrical input energy with the thermal load.
The model uses 4.186 kJ/(kg K) for liquid water, 2.1 for ice and 333.55 kJ/kg for fusion. These are fixed approximations, not temperature-dependent property tables or corrections for dissolved ingredients.
Interpret the result
Duration divides required heat by a constant useful cooling rate. Container mass, ambient heat leak, cycling, declining capacity and ice nucleation can alter the actual time even if the total ideal energy balance is correct.
Input energy is thermal energy divided by entered COP. Multiplying time by thermal capacity and then treating it as electrical power would confuse two different sides of the cooling process.
Work through the example
One kilogram starting at twenty degrees needs 83.72 kJ removed to reach zero. Freezing removes 333.55 kJ, and cooling the ice to minus ten removes another twenty-one.
Total removal is 438.27 kJ, or about 0.121742 thermal kWh. At 100 useful thermal watts the ideal duration is about 1.21742 hours; COP two gives about 0.060871 electrical kWh under the entered assumptions.
Where the model stops
Pure-water constant-property ideal model only; no container, heat ingress, nucleation, pressure effect or equipment selection.
Real freezing duration and appliance input require actual operating data. This page does not recommend freezer sizing or food freezing procedures.
A few questions, answered
Why is there an energy term with no temperature change?
Latent heat changes the phase from liquid to solid at the assumed melting boundary.
Can I use the freezer’s electrical watts as cooling capacity?
Not directly. Useful thermal removal and electrical input are related by COP and operating conditions.