Science, Tech, Math › Science Calculate Energy Required to Turn Ice Into Steam Heat Calculation Example Problem Share Flipboard Email Print Ice undergoes phase changes to become steam. Left: Atomic Imagery/Getty Images; Right: sandsun/Getty Images Science Chemistry Basics Chemical Laws Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Todd Helmenstine Todd Helmenstine is a science writer and illustrator who has taught physics and math at the college level. He holds bachelor's degrees in both physics and mathematics. our editorial process Todd Helmenstine Updated November 05, 2018 This worked example problem demonstrates how to calculate the energy required to raise the temperature of a sample that includes changes in phase. This problem finds the energy required to turn cold ice into hot steam. Ice to Steam Energy Problem What is the heat in Joules required to convert 25 grams of -10 °C ice into 150 °C steam?Useful information:heat of fusion of water = 334 J/gheat of vaporization of water = 2257 J/gspecific heat of ice = 2.09 J/g·°Cspecific heat of water = 4.18 J/g·°Cspecific heat of steam = 2.09 J/g·°CSolution:The total energy required is the sum of the energy to heat the -10 °C ice to 0 °C ice, melting the 0 °C ice into 0 °C water, heating the water to 100 °C, converting 100 °C water to 100 °C steam and heating the steam to 150 °C. To get the final value, first calculate the individual energy values and then add them up.Step 1: Heat required to raise the temperature of ice from -10 °C to 0 °C Use the formulaq = mcΔTwhereq = heat energym = massc = specific heatΔT = change in temperatureq = (25 g)x(2.09 J/g·°C)[(0 °C - -10 °C)]q = (25 g)x(2.09 J/g·°C)x(10 °C)q = 522.5 JHeat required to raise the temperature of ice from -10 °C to 0 °C = 522.5 JStep 2: Heat required to convert 0 °C ice to 0 °C waterUse the formula for heat:q = m·ΔHfwhereq = heat energym = massΔHf = heat of fusionq = (25 g)x(334 J/g)q = 8350 JHeat required to convert 0 °C ice to 0 °C water = 8350 JStep 3: Heat required to raise the temperature of 0 °C water to 100 °C waterq = mcΔTq = (25 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]q = (25 g)x(4.18 J/g·°C)x(100 °C)q = 10450 JHeat required to raise the temperature of 0 °C water to 100 °C water = 10450 JStep 4: Heat required to convert 100 °C water to 100 °C steamq = m·ΔHvwhereq = heat energym = massΔHv = heat of vaporizationq = (25 g)x(2257 J/g)q = 56425 JHeat required to convert 100 °C water to 100 °C steam = 56425Step 5: Heat required to convert 100 °C steam to 150 °C steamq = mcΔTq = (25 g)x(2.09 J/g·°C)[(150 °C - 100 °C)]q = (25 g)x(2.09 J/g·°C)x(50 °C)q = 2612.5 JHeat required to convert 100 °C steam to 150 °C steam = 2612.5Step 6: Find total heat energyHeatTotal = HeatStep 1 + HeatStep 2 + HeatStep 3 + HeatStep 4 + HeatStep 5HeatTotal = 522.5 J + 8350 J + 10450 J + 56425 J + 2612.5 JHeatTotal = 78360 JAnswer:The heat required to convert 25 grams of -10 °C ice into 150 °C steam is 78360 J or 78.36 kJ.