Question

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

Answer 1

The scenario has to be divided into 5 steps

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

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

∆H_vap = Heat of vaporization at 100^oC = 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