Answer:
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Heat to be withdrawn to cool water from 22.4 degree to 0 degree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Heat to be withdrawn to cool water 0 degree ice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Heat to be withdrawn to cool ice from 0 degree to -7.2 degree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Total heat to be withdrawn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
