Question

. Calculate the total heat required (a) to melt 180 g of ice at 0 0 C, (b) heat it to 100 0 C and then (c) vapourise it at that temperature. Given ∆fusH0 (ice) = 6.01 kJ mol-1 at 0 0 C, ∆ vap H0 (H2 O) = 40.7 kJ mol-1 at 100 0 C specific heat of water is 4.18 J g-1 K-1

Answer 1

Given:

∆fus H (ice) = 6.01 kJ/mole

∆vap H (H2O) = 40.7 kJ/mole

The specific heat of water, c = 4.18 J /gK

To Find:

Heat Required:

(a) to melt 180 g of ice at 0 0 C,

(b) to heat it to 100 0 C

(c) to vapourise it at that temperature.

Calculation:

(a) Heat required to melt 180 g of ice at 0 C:

Q = m ∆fus H (ice)

⇒ Q = 180 × 6.01 = 1081.8 kJ

(b) Heat required to heat it to 100 C:

Q = mcΔT

⇒ Q = 180 × 4.18 × 100 = 75.24 kJ

(c) Heat required to vapourise it:

Q = m ∆vap H (H2O)

⇒ Q = 180 × 40.7 = 7326 kJ

- So, the total heat required:

(a) to melt 180 g of ice at 0 C = 1081.8 kJ

(b) to heat it to 100 C = 75.24 kJ

(c) to vapourise it = 7326 kJ