Question

# What amount of thermal energy (in kJ) is required to convert 220 g of ice at...

What amount of thermal energy (in kJ) is required to convert 220 g of ice at -18 °C completely to water vapour at 248 °C? The melting point of water is 0 °C and its normal boiling point is 100 °C. The heat of fusion of water is 6.02 kJ mol-1 The heat of vaporization of water at its normal boiling point is 40.7 kJ mol-1 The specific heat capacity of ice is 2.09 J g-1 °C-1 The specific heat capacity of liquid water is 4.18 J g-1 °C-1 The specific heat capacity of water vapour is 2.01 J g-1 °C-1

The scenario has to be divided into 5 steps

1. H2O (s) at -18oC to H2O (s) at 0oC (Thermal energy required, q1)
2. H2O (s) at 0oC to H2O (l) at 0oC (Thermal energy required, q2)
3. H2O (l) at 0oC to H2O (l) at 100oC (Thermal energy required, q3)
4. H2O (l) at 100oC to H2O (g) at 100oC (Thermal energy required, q4)
5. H2O (g) at 100oC to H2O (g) at 248oC (Thermal energy required, q5)

Total thermal energy required = Q = q1 +q2 + q3 + q4 + q5

[Given date and notations,

Mass of the ice taken, m = 220 g

∆T = represents the temperature change in oC

∆Hfus = Heat of fusion of water = 6.02 kJ/mol = 0.334 kJ/g

∆Hvap = Heat of vaporization at 100oC = 40.7 kJ/mol = 2.261 kJ/g

Specific heat capacity = ‘c’]

Total thermal energy required = Q = q1 +q2 + q3 + q4 + q5

Q = 8.2764 + 73.5700 + 91.9600 + 497.4400 + 65.4456 = 736.6920 kJ

#### Earn Coins

Coins can be redeemed for fabulous gifts.