Question

At 1 atm, how much energy is required to heat 81.0 g of H2O(s) at –20.0 °C to H2O(g) at 157.0 °C? Helpful constants can be found here.

Answer 1

Mass of water = 81.0 g

Enthalpy of fusion of ice = 333.55 J/g

Enthalpy of vaporisation of water = 2260 J/g

Heat capacity of water = 4.184 J/g

Heat required to heat 81.0 g of H2O(s) at –20.0 °C to H2O(g) at 157.0 °C

(81.0gx20x4.184J/g) + (80.0 gx333.55 J/g) + (81.0gx100x4.184J/g) + (80.0 gx2260 J/g) + (81.0gx57x4.184J/g)

= 6778.08 J + 26684 J + 33890.4 J +180800 J +19317.5 J

= 267469.98 J

= 267.69 kJ

The amount of energy required to heat 81.0 g of H2O(s) at –20.0 °C to H2O(g) at 157.0 °C is 267.69 kJ