Converting 23.0 g of ice at -10.0 °C to steam at 109 °C requires 69.8 kJ of energy.
There are five heats to consider:
#q_1# = heat required to warm the ice to 0.00 °C. #q_2# = heat required to melt the ice to water at 0.00 °C. #q_3# = heat required to warm the water from 0.00 °C to 100.00 °C. #q_4# = heat required to vapourize the water to vapour at 100 °C. #q_5# = heat required to warm the vapour to 109 °C.
#q_1 = mcΔT# = 23.0 g × 2.108 J•°C⁻¹g⁻¹ × 10.0 °C = 484.84 J
#q_2 = mΔH_"fus"# = 23.0 g × 334 J• g⁻¹ = 7682 J
#q_3 = mcΔT# = 23.0 g × 4.184 J°C⁻¹g⁻¹ × 100.00 °C = 9623.2 J
