# How much energy is required to convert 100.g of ice at 0.00 °C to water vapor at 100.00 °C?

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Feb 28, 2014

Converting 100. g of ice at 0.00 °C to water vapour at 100.00 °C requires 301 kJ of energy.

There are three heats to consider:

${q}_{1}$ = heat required to melt the ice to water at 0.00 °C.

${q}_{2}$ = heat required to warm the water from 0.00 °C to 100.00 °C.

${q}_{3}$ = heat required to vapourize the water to vapour at 100 °C.

q_1 = mΔH_(fus) = 100. g × (334 J)/(1 g) = 33 400 J

q_2 = mcΔT = 100. g × (4.184 J)/(1 °C•g) × 100.00 °C = 41 800 J

q_3 = mΔH_(vap) = 100. g × (2260 J)/(1 g) = 226 000 J

${q}_{1} + {q}_{2} + {q}_{3}$ = 33 400 J + 41 800 J + 226 000 J = 301 000 J
= 301 kJ

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