Transcript Lecture 25 - Purdue University

PHYSICS 220
Lecture 24
Heat
Lecture 24
Purdue University, Physics 220
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Internal Energy
• Energy of all molecules including
– Translational and rotational kinetic energy of molecules due to
their individual random motions.
– Vibrational energy (both kinetic and potential) of molecules and of
atoms within molecules due to random vibrations about their
equilibrium points.
– Potential energy due to interactions between the atoms and
molecules of the systems.
– Chemical and nuclear energy (kinetic and potential energy
associated with the binding of atoms to form molecules, the
binding of electrons to nuclei to form atoms, and the binding of
protons and neutrons to form nuclei).
• DOES NOT INCLUDE
– Macroscopic motion of object
– Potential energy due to interactions with other objects
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Heat
• Definition: Flow of energy between two objects
due to difference in temperature
– Note: similar to WORK
– Object does not “have” heat (it has energy)
• Units: calorie
– Amount of heat needed to raise 1g of water 1ºC
– 1 Calorie = 1 kcal = 1000 cal = 4186 Joules
• Heat flows from a system at higher temperature
to one at lower temperature
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Heat Capacity and Specific Heat
Heat capacity =Q/T
• shows how much heat is required to
change the T of object (system)
• Specific heat c= Q/mT
• Q = c m T
Heat required to increase temperature
depends on amount of material (m) and type
of material
• Heat adds energy to object/system
• IF there is no dissipation then:
Heat increases internal energy: Q = U
Heat increases temperature: Q = C T
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Exercise
After a grueling work out, you drink a liter of cold
water (0 C). How many Calories does it take for
your body to raise the water up to body
temperature of 36 C?
A) 36
B) 360
C) 3,600
D) 36,000
1 liter = 1,000 grams of H20
1000 g x 1 calorie/(gram degree) x (36 degree) = 36,000 calories
36,000 calories = 36 Calories!
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Question
Suppose you have equal masses of aluminum and
copper at the same initial temperature. You add 1000 J
of heat to each of them. Which one ends up at the
higher final temperature
A) aluminum
B) copper
C) the same
Substance
aluminum
copper
iron
lead
human body
water
ice
c in J/(kg-C)
900
387
452
128
3500
4186
2000
T = Q/cm
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iClicker
Two insulated buckets contain the same amount of water at
room temperature. Two blocks of metal of the same mass,
both at the same temperature, are warmer than the water in
the buckets. One block is made of aluminum (c=0.9) and one
is made of copper. You put the aluminum block into one
bucket of water, and the copper (c=0.385) block into the other.
After waiting a while you measure the temperature of the
water in both buckets. Which is warmer?
A) The water in the bucket containing the aluminum block
B) The water in the bucket containing the copper block
C) The water in both buckets will be at the same temperature
Since aluminum has a higher specific heat than copper, you are
adding more heat to the water when you dump the aluminum in the
bucket (Q=mcT).
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Specific Heat for Ideal Gas
• Monatomic Gas (single atom)
Translational kinetic energy only
At constant Volume work = 0
Q = U = 3/2 nRT
CV = 3/2 R = 12.5 J/(K mole)
Cv – specific heat at constant volume.
• Diatomic Gas (two atoms)
Can also rotate
CV = 5/2 R = 20.8 J/(K mole)
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Phase Transitions
• A phase transition occurs whenever a material is changed
from one phase, such as the solid phase, to another
phase, such as the liquid phase.
– Phase transitions occur at constant temperature.
– The latent heat of vaporization LV is the heat per unit mass that
must flow to change the phase from liquid to gas or from gas to
liquid.
• Fusion occurs when a liquid turns into a solid.
• Evaporation occurs when a liquid turns into a gas.
• Sublimation occurs when a solid changes directly to a gas
without going into a liquid form.
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Latent Heat
• As you add heat to water, the temperature increases
for a while, then it remains constant, despite the
additional heat!
steam
T
Substance
water
Lf (J/kg)
33.5 x 104
f=fusion
100oC
Lv (J/kg)
22.6 x 105
water
temp
rises
v=vaporization
water
changes
to steam
(boils)
temp
rises
Latent Heat
Q added to water
• Latent Heat L [J/kg] is heat which must be added (or
removed) for material to change phase (liquid-gas).
• |Q| = m L
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Phase Diagram
H2O
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Phase Diagram
CO2
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iClicker
Which can absorb more energy from your soda, a
“cooler” filled with water at 0 C, or a cooler filled
with ice at 0 C.
A) Water
B) About Same
C) Ice
Latent Heat L [J/kg] is heat which must be added
(or removed) for material to change phase (liquidwater
ice
gas).
ice
T
Substance
water
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Lf (J/kg)
33.5 x 104
0oC
Lv (J/kg)
22.6 x 105
temp
rises
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changes
to water
(melts)
temp
rises
Latent Heat
Q added to water
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Exercise
During a tough work out, your body sweats (and evaporates)
1 liter of water to keep cool (37 C). How much cold water
would you need to drink (at 2 C) to achieve the same thermal
cooling? (recall CV = 4.2 J/g for water, Lv=2.2x103 J/g)
A) 0.15 liters
B) 1.0 liters
C) 15 liters
D) 150 liters
Qevaporative = L m = 2.2x103 kJ/kg x 1kg
Qc = c m t = 4.2kJ/kgK x 35K x m
m = 2.2x103 / 147 = 15kg or 15 liters!
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Boiling Point
Going from Lafayette to Denver the
temperature at which water boils:
A) Increases
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B) Decreases
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C) Same
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Exercise
How much ice (at 0 C) do you need to add to 0.5 liters of a
water at 25 C, to cool it down to 10 C?
(L = 80 cal/g, c = 1 cal/g C)
Qwater  mcT
 (0.5kg)(1cal / gC)(15C)
 (7,500 calories)
Qice  mL  mcT
 m
Qice
L  cT
7,500cal

 83.3 grams
80cal / g  (1cal / gC)(10)
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Exercise
Ice cube trays are filled with 0.5 kg of water at 20 C and
placed into the freezer. How much energy must be
removed from the water to turn it into ice cubes at -5 C?
(L = 80 cal/g, cwater = 1 cal/g C, cice = 0.5 cal/g C)
Water going from 20 C to 0 C:
Q1  mcwater T1
 500  1 (20)  10000(cal)
Water turning into ice at 0 C:
Q2  mL
 500  80  40000(cal)
Ice going from 0 C to -5 C:
Q3  mcice T2
 500  0.5  (5)  1250(cal)
 Q  Q1  Q2  Q3  51250(cal)
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