Question

Calculate the total heat required 

(a) to melt 180 g of ice at 0°C, 

(b) heat it to 100°C and then 

(c) vapourise it at that temperature.

Given ∆_fusH_(ice) = 6.01 kJ mol^-1 at 0°C,

∆_vapH_(H2O) = 40.7 kJ mol^-1 at 100°C specific heat of water is 4.18 J g^-1 K^-1.

Answer 1

Given : Mass of ice = m = 180 g

Number of moles of H₂O = \(\frac{180}{18}\) = 10 mol

The total process can be represented as,

(i) ∆H₁ = ∆_fusH = 10 mol × 6.01 kJ mol^-1 = 60.1 kJ mol^-1

(ii) When the temperature of water is raised from 0°C to 100°C (i.e., 273 K to 373 K), then

Hence total enthalpy change,