One gram of steam at 100 degrees C condenses and the water cools to 22 degrees C

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.

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·°Cspecific heat of water = 4.18 J/g·°Cspecific heat of steam = 2.09 J/g·°C

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

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

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 = mcΔTq = (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 JThe 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·ΔHvwhere

q = heat energy

ΔHv = heat of vaporization

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 steamq = mcΔTq = (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

Answer:

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

• Atkins, Peter and Loretta Jones (2008). 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.
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