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Calculate Energy Required To Turn Ice into Steam

Heat Calculation Example Problem

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Ice undergoes phase changes to become steam.

Ice undergoes phase changes to become steam.

Anthony Bradshaw, Getty Images
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.

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/g
heat of vaporization of water = 2257 J/g
specific heat of ice = 2.09 J/g·°C
specific heat of water = 4.18 J/g·°C
specific heat of steam = 2.09 J/g·°C

Solution:

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.

Step 1: Heat required to raise the temperature of ice from -10 °C to 0 °C Use the formula

q = mcΔT

where
q = heat energy
m = mass
c = specific heat
ΔT = change in temperature

q = (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 J

Heat required to raise the temperature of ice from -10 °C to 0 °C = 522.5 J

Step 2: Heat required to convert 0 °C ice to 0 °C water

Use the formula

q = m·ΔHf

where
q = heat energy
m = mass
ΔHf = heat of fusion

q = (25 g)x(334 J/g)
q = 8350 J

Heat required to convert 0 °C ice to 0 °C water = 8350 J

Step 3: Heat required to raise the temperature of 0 °C water to 100 °C water

q = mcΔT

q = (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 J

Heat required to raise the temperature of 0 °C water to 100 °C water = 10450 J

Step 4: Heat required to convert 100 °C water to 100 °C steam

q = m·ΔHv

where
q = heat energy
m = mass
ΔHv = heat of vaporization

q = (25 g)x(2257 J/g)
q = 56425 J

Heat required to convert 100 °C water to 100 °C steam = 56425

Step 5: Heat required to convert 100 °C steam to 150 °C steam

q = mcΔT
q = (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 J

Heat required to convert 100 °C steam to 150 °C steam = 2612.5

Step 6: Find total heat energy

HeatTotal = HeatStep 1 + HeatStep 2 + HeatStep 3 + HeatStep 4 + HeatStep 5
HeatTotal = 522.5 J + 8350 J + 10450 J + 56425 J + 2612.5 J
HeatTotal = 78360 J

Answer:

The heat required to convert 25 grams of -10 °C ice into 150 °C steam is 78360 J or 78.36 kJ.

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