# At what rate does heat have to be removed in the following case:A cold storage has 1500 kg of water at 20 degree Celsius that has to be converted to ice at 0 degree Celsius. The ice has to be...

At what rate does heat have to be removed in the following case:

A cold storage has 1500 kg of water at 20 degree Celsius that has to be converted to ice at 0 degree Celsius. The ice has to be created within half an hour.

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The cold storage has 1500 kg of water at 20 degree Celsius. This has to be converted to ice at 0 degree Celsius in half an hour.

First we have to cool water to 0 degree Celsius. The amount of heat to be removed to do this is given by Cp*mass*temperature, where Cp is the heat capacity of water given as 4.1813 kJ/kg*K. For 1500 kg the heat to be removed is equal to 1500*4.1813*20 = 125439 kJ

Once the water has been cooled to 0 degree Celsius we are not done, now heat has to be removed to convert it to ice. This is equal to the specific heat of fusion of water or 333.55 kJ/kg. For 1500 kg it is 1500*333.55 = 500325 kJ.

This gives the total heat that has to be removed from the water as 125439 + 500325 = 625764 kJ

As the ice has to be created in 30 minutes, the rate of cooling required is 625764/30*60 = 347.64 kJ/s or 347.64 kW.

The required rate at which heat has to be removed is 347.64 kW.