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_1q1 = heat required to warm the ice to 0.00 °C. q_2q2 = heat required to melt the ice to water at 0.00 °C. q_3q3 = heat required to warm the water from 0.00 °C to 100.00 °C. q_4q4 = heat required to vapourize the water to vapour at 100 °C. q_5q5 = 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
