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: Δ_fusH° (ice) = 6.01 kJ mol^-1 at 0 °C, Δ_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: 

Δ_fusH° (ice) = 6.01 kJ mol^-1 at 0 °C,
Δ_vapH° (H₂O) = 40.7 kJ mol^-1 at 100 °C,
Specific heat of water is 4.18 J g^-1 K^-1

To find:

The total heat required to carry out the given reaction using 180 g of ice.

Calculation:

\[\ce{\underset{\text{(ice at 0 °C)}}{H2O_{(s)}} ->[Latent heat][of fusion 0 °C]\underset{\text{(water at 0 °C)}}{H2O_{(l)}}->[Heating]\underset{\text{(water at 100 °C)}}{H2O_{(l)}} ->[Latent][haeat of vaporization 100 °C] \underset{\text{(Stream at 100 °C)}}{H2O_{(g)}}}\]

a) H₂O_(s) → H₂O_(l)
    0 °C          0 °C
Heat required = Latent heat for 180 g.

1 mol of H₂O = 6.01 kJ

1 mol of H₂O = 18 g

∴ 180 g of H2O = `(180 "g")/(18 "g mol"^-1)` = 10 moles of H₂O

∴ 10 mol of H₂O requires = 60.1 kJ

∴ Heat required = 60.1 kJ …(i)

b) H₂O_(l) → H₂O_(s)
    0 °C          100 °C

Heat required = Mass × Specific heat × ΔT

= 180 g × 4.18 J g^-1 K^-1 × 100 K

= 75240 J

= 75.240 kJ    ....(ii) 

c) H₂O_(l) → H₂O_(g)
100 °C        100 °C

Heat required = Latent heat of vaporization

1 mol of H₂O requires = 40.7 kJ

∴ 1 mol of H₂O = 18 g

∴ 180 g of H₂O = 10 moles of H₂O

∴ Heat required by 10 moles of water = 407 kJ ….(iii)

From (i) , (ii) and (iii),

Total heat required to carry out the given reaction using 180 g of ice

= 60.1 kJ + 75.240 kJ + 407 kJ = + 542.34 kJ

The total heat required to melt 180 g of ice at 0 °C, heat it to 100 °C and then vaporize it at that temperature is + 542.34 kJ.