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·ΔH_{f}

where

- q = heat energy
- m = mass
- ΔH
_{f}= heat of fusion

For this problem:

- q = ?
- m = 25 g
- ΔH
_{f}= 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·ΔH_{v}

where

q = heat energy

m = mass

ΔH_{v} = 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.

Heat_{Total} = Heat_{Step 1} + Heat_{Step 2} + Heat_{Step 3} + Heat_{Step 4} + Heat_{Step 5}

Heat_{Total} = 522.5 J + 8350 J + 10450 J + 56425 J + 2612.5 J

Heat_{Total} = 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.

## Sources

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*Chemical Principles: The Quest for Insight*(4th ed.). W. H. Freeman and Company. p. 236. ISBN 0-7167-7355-4. - Ge, Xinlei; Wang, Xidong (2009). "Calculations of Freezing Point Depression, Boiling Point Elevation, Vapor Pressure and Enthalpies of Vaporization of Electrolyte Solutions by a Modified Three-Characteristic Parameter Correlation Model".
*Journal of Solution Chemistry*. 38 (9): 1097–1117. doi:10.1007/s10953-009-9433-0 - Ott, B.J. Bevan and Juliana Boerio-Goates (2000)
*Chemical Thermodynamics: Advanced Applications*. Academic Press. ISBN 0-12-530985-6. - Young, Francis W.; Sears, Mark W.; Zemansky, Hugh D. (1982).
*University Physics*(6th ed.). Reading, Mass.: Addison-Wesley. ISBN 978-0-201-07199-3.