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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. - Chemistry

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Sum

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]

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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

 

Concept: Thermochemistry
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APPEARS IN

Balbharati Chemistry 12th Standard HSC for Maharashtra State Board
Chapter 4 Chemical Thermodynamics
Exercises | Q 4.18 | Page 89
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