# How much energy is needed to convert 23.0 grams of ice at -10.0°C into steam at 109°C?

Jul 25, 2014

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

q_4 = mΔH_"vap" = 23.0 g × 2260 J•g⁻¹ = 51 980 J

q_5 = mcΔT = 23.0 g ×2.01 J•g⁻¹ = 46.23 J

${q}_{1} + {q}_{2} + {q}_{3} + {q}_{4} + {q}_{5}$ =( 484.84 + 7682 + 9623.2 + 51 980 + 46.23) J = "69 816.27" J = 69.8 kJ (3 significant figures)