What is the total number of joules required to freeze a 10 g sample of water at 0c

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.
• Young, Francis W.; Sears, Mark W.; Zemansky, Hugh D. (1982). University Physics (6th ed.). Reading, Mass.: Addison-Wesley. ISBN 978-0-201-07199-3.

Heat of Fusion-the amount of heat required to convert unit mass of a solid into the liquid without a change in temperature. (or released for freezing)

For water at its normal freezing point of 0 ºC, the specific heat of Fusion is 334 J g-1. This means that to convert 1 g of ice at 0 ºC to 1 g of  water at 0 ºC, 334 J of heat must be absorbed by the water. Conversely, when 1 g of water at 0 ºC freezes to give 1 g of ice  at 0 ºC, 334 J of heat will be released to the surroundings.

Heat of Fusion of Water (Hf = 334 J /g)

q= m Hf

Note- The Heat of Fusion equation is used only at the melting/freezing transition, where the temperature remains the same only and that is why there is no temperature change (DT) in this formula. It stays at 0 Celsius for water.

Note #2-Energy is required to melt and released when it freezes

The diagram on the left shows the uptake of heat by 1 kg of water, as it passes from ice at -50 ºC to steam at temperatures above 100 ºC, affects the temperature of the sample. E: Steam absorbs heat and thus increases its temperature. D: Water boils and absorbs latent heat of vaporization. C: Rise in temperature as liquid water absorbs heat. B: Absorption of latent heat of fusion. A: Rise in temperature as ice absorbs heat. from-http://www.physchem.co.za/Heat/Latent.htm

The first important thing to mention here is that you don't need heat to freeze water at its freezing point!

In fact, heat is being given off when water goes from liquid at #0^@"C"# to solid at #0^@"C"#, i.e. when it undergoes a solid #-># liquid phase change.

In other words, freezing is an exothermic process because the system is giving off heat to its surroundings.

Now, the enthalpy of fusion, #DeltaH_f#, tells you how much heat is needed in order to convert ice at #0^@"C"# to liquid water at #0^@"C"#. For water, the enthalpy of fusion is given as

#DeltaH_f = "333.55 J g"^(-1)#

The trick here is to realize that the amount of heat needed to melt ice will be equal to the amount of heat given off when liquid water freezes.

You can thus say that when #"1 g"# of water freezes at #0^@"C"#, #"333.55 J"# of heat are being given off to the surroundings. Since your sample has a mass of #"295 g"#, it follows that it will release

#295 color(red)(cancel(color(black)("g"))) * "333.55 J"/(1color(red)(cancel(color(black)("g")))) = "98,397 J"#

Rounded to three sig figs and expressed in kilojoules, the answer will be

#"heat given off" = color(green)(|bar(ul(color(white)(a/a)color(black)("98.4 kJ")color(white)(a/a)|)))#