A quantity of 30.1 cm^3 of water at 22.4°C is placed in a freezer compartment and allowed to freeze to solid ice at - 7.2°C. How many joules of energy must be withdrawn from the water by the refrigerator?

13330.86 J

Explanation:

mass of volume of 30.1 cc of water = density x volume

= 1 x 30.1 = 30.1 gm

= 30.1 x 10⁻³ kg

Heat to be withdrawn to cool water from 22.4 degree to 0 degree

= mass x specific heat x fall of temperature

= 30.1 x 10⁻³ x 4186 x 22.4

= 2822.36 J

Heat to be withdrawn to cool water 0 degree ice

mass x latent heat of freezing

30.1 x 10⁻³ x 334000

= 10053.4 J

Heat to be withdrawn to cool ice from 0 degree to - 7.2 degree

mass x specific heat of ice x fall of temperature

= 30.1 x 10⁻³ x 2100 x 7.2

= 455.1 J

Total heat to be withdrawn

= 2822.36 + 10053.4 + 455.1

= 13330.86 J