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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: ΔfusH° (ice) = 6.01 kJ mol-1 at 0 °C, ΔvapH° (H2O) = 40.7 kJ mol-1 at 100 °C, Specific heat of water is 4.18 J g-1 K-1]
Solution
Given:
ΔfusH° (ice) = 6.01 kJ mol-1 at 0 °C,
ΔvapH° (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.
Notes
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