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# 12.8 Heat and Phase Change: Latent Heat

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12.8 Heat and Phase Change: Latent Heat
Glass of ice water at 0 C. Heat is being used to melt ice, and then only when all the ice is melted will the temperature of the liquid begin to rise.

Heat and Phase Change: Latent Heat
A solid can melt or fuse into a liquid if heat is added. Liquid can freeze into a solid if heat is removed. A liquid can evaporate into a gas if heat is supplied. Gas can condense into a liquid if heat is taken away. Rapid evaporation, formation of vapor bubbles within the liquid is called boiling. A solid can change directly into a gas if that is provided, called sublimation.

Example: dry ice (CO2) Solid naphthalene (moth balls) turns into naphthalene fumes.

Conceptual Example 13: Saving Energy
Suppose you are cooking spaghetti for dinner, and the instructions say “boil the pasta in water for ten minutes.” To cook spaghetti in an open pot with the least amount of energy, should you turn up the burner to its fullest so the water vigorously boils, or should you turn down the burner so the water barely boils?

Latent Heat Q = mL Heat needed for a phase change.
A substance changes from one phase to another, the amount of heat that must be added or removed depends on the type of material and the nature of the phase change. The heat Q that must be supplied or removed to change the phase of a mass m of a substance is. Q = mL Where L is the latent heat of the substance. SI Unit of Latent Heat: j/kg

Latent Heat Latent heat of fusion: Lf, refers to the change between solid and liquid phases. Latent heat of vaporization Lv, change between liquid and gas phases. Latent heat of sublimation Ls: change between solid and gas phases. Lf = 3.35 x 10^5 J/kg water So 3.35 x 10^5 J/kg of heat must be supplied to melt one kilogram of ice at 0 C. This amt. of heat must be removed from one kilogram of liquid water at 0 C to freeze the liquid into ice.

Latent heat of vaporization for water has the much larger value of Lv = 22.6 x 10^5 J/kg
When water boils at 100 C, 22.6 x 10^5 J of heat must be supplied for each kilogram of liquid turned into steam. When steam condenses at 100 C, this amt. of heat is released from each kilogram of steam that changes back into liquid.

In the Real World Designers can engineer clothing that can absorb or release heat to help maintain a comfortable and constant temperature close to your body. PCM “phase change material” Prevents overheating by melting, absorbing excess body heat in the process. PCM freezes and releases heat to keep you warm.

Example 14: Ice-Cold Lemonade
Ice at 0 C is placed in a styrofoam cup containing 0.32 kg of lemonade at 27 C. The specific heat capacity of lemonade is virtually the same as that of water; that is, c = 4186 J/(kgxC). After the ice and lmonade reach an equilibrium temperature, some ice still remains. The latent heat of fusion for water is Lf = 3.35 x 10^5 J/kg. Assume that the mass of the cup is so small that it absorbs a negligible amount of heat, and ignore any heat lost to the surroundings. Determine the mass of ice that has melted.

Example 15: Getting Ready for a Party
A 7.00kg glass bowl (c = 840 J/(kgxC)) contains 16.0kg of punch at 25 C. Two and a half kilograms of ice [c=2.00 x 10^3J/(kgxC0] are added to the punch. The ice has an initial temperature of C, having been kept in a very cold freezer. The punch may be treated as if it were water [c = 4186J/(kgxC)], and it may be assumed that there is no heat flow between the punch bowl and the external environment. The latent heat of fusion for water is 3.35 x 10^5 J/kg. When thermal equilibrium is reached, all the ice has melted, and the final temperature of the mixture is above 0 C. Determine this temperature.

Practice Problem 52. To help prevent frost damage, fruit growers sometimes protect their crop by spraying it with water when overnight temperatures are expected to go below the freezing mark. When the water turns to ice during the night, that is released into the plants, thereby giving them a measure of protection against the falling temperature. Suppose a grower sprays 7.2 kg of water at 0 C onto a fruit tree. (a) How much heat is released by the water when it freezes? (b) How much would the temperature of a 180 kg tree rise if it absorbed the heat released in part (a)? Assume that the specific heat capacity of the tree is 2.5 x 10^3 J/kgC and that no phase change occurs within the tree itself.

Homework

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