Question

How much heat (in kJ) is required to warm 10.0 g of ice, initially at -15.0 ∘C, to steam at 111.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C.

Answer 1

We need the heat reuired in (KJ), so convert the heat capacities required in the process also into KJ

specific heat of water is 4.186 kJ/kgC

specific heat of ice is 2.09 kJ/kgC
specific heat of steam is 2.01 kJ/kgK
heat of fusion of ice is 334 kJ/kg
heat of vaporization of water is 2256 kJ/kg

5 parts to this problem, add them up to get the total in kJ
10g = 0.01 kg

to warm ice from -15C to 0C
E1 = 2.09 kJ/kgC x 0.01 kg x (0-(-15))C = 0.3135 kJ

to melt ice
E2 = 334 kJ/kg x 0.01 kg = 3.34 kJ

to warm water from 0C to 100C
E3 = 4.186 kJ/kgC x 0.01 kg x (100-0)C = 4.186 kJ

to boil water
E4 = 2256 kJ/kg x 0.01 kg = 22.56 kJ

to heat stem from 100C to 111C
E5 = 2.01 kJ/kgC x 0.01 kg x (111-100)C = 0.2211 kJ

Total heat required for the process in (kJ) = E1 +E2+E3+E4+E5

                                          = 0.3135+3.34+4.186+22.56+0.2211 = 30.62 KJ