Calculate Energy Required to Turn Ice Into Steam

Calculate Energy Required to Turn Ice Into Steam 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 ·Ã‚ °Cspecific heat of water 4.18 J/g ·Ã‚ °Cspecific heat of steam 2.09 J/g ·Ã‚ °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 ·Ã‚ °C)[(0  °C - -10  °C)]q (25 g)x(2.09 J/g ·Ã‚ °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 ·ÃŽâ€Hfwhereq 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 ·Ã‚ °C)[(100  °C - 0  °C)]q (25 g)x(4.18 J/g ·Ã‚ °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 ·ÃŽâ€Hvwhereq 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 ·Ã‚ °C)[(150  °C - 100  °C)]q (25 g)x(2.09 J/g ·Ã‚ °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.