chapt 12 part2 - Physics-YISS

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Transcript chapt 12 part2 - Physics-YISS

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
• Liquid can freeze into a solid if heat is removed.
• A liquid can evaporate into a gas if heat is
• Gas can condense into a liquid if heat is taken
• 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 (CO )
• 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
• 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 -20.0 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.