# Calculate Energy Required to Turn Ice Into Steam

## Heat Calculation Example Problem Ice undergoes phase changes to become steam. Left: Atomic Imagery/Getty Images; Right: sandsun/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.

## 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/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

## Solving the Problem

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:

Find the 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

In this problem:

• q = ?
• m = 25 g
• c = (2.09 J/g·°C
• ΔT = 0 °C - -10 °C (Remember, when you subtract a negative number, it is the same as adding a positive number.)

Plug in the values and solve for q:

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

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

Step 2:

Find the heat required to convert 0 °C ice to 0 °C water.

Use the formula for heat:

q = m·ΔHf

where

For this problem:

• q = ?
• m = 25 g
• ΔHf = 334 J/g

Plugging in the values gives the value for q:

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

The heat required to convert 0 °C ice to 0 °C water = 8350 J

Step 3:

Find the 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
The heat required to raise the temperature of 0 °C water to 100 °C water = 10450 J
Step 4:

Find the 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
The heat required to convert 100 °C water to 100 °C steam = 56425

Step 5:

Find the 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
The heat required to convert 100 °C steam to 150 °C steam = 2612.5

Step 6:

Find total heat energy. In this final step, put together all of the answers from the previous calculations to cover the entire temperature range.

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