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**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. 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 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 for heat:

q = m·ΔH_{f}

where

q = heat energy

m = mass

ΔH_{f} = 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·ΔH_{v}

where

q = heat energy

m = mass

ΔH_{v} = 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

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.