Physics 231 Topic 13: Heat Wade Fisher Nov 19-28 2012

Physics 231
Topic 13: Heat
Wade Fisher
Nov
19-28
2012
MSU Physics
231 Fall
2012
1
Clicker Quiz!
What happens to the volume
of a balloon if you put it in the
a) it increases
b) it does not change
c) it decreases
freezer?
According to the Ideal Gas Law, when the temperature is reduced at
constant pressure, the volume is reduced as well. The volume of the
balloon therefore decreases.
PV = nRT
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Key Concepts: Heat
Heat and Thermal Energy
Heat & its units
Heat Capacity & Specific Heat
Thermal Equilibrium
Equipartition theorem
Phase Changes
Latent heat of fusion, vaporization
Conduction, Convection, and Radiation
Thermal Conductivity
Stefan-Boltzman law
Covers chapter 13 in Rex & Wolfson
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Internal Energy
In chapter 12: The internal (total) energy for an ideal
gas is the total kinetic energy of the atoms/particles
in a gas.
For a non-ideal gas: the internal energy is due to kinetic
and potential energy associated with:
• translational motion
• rotational motion
• vibrational motion
• intermolecular potential energy
|PEideal gas=0| < |PEnon-ideal gas| < |PEliquid| < |PEsolid|
PE
PE: negative!
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R
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Heat
Heat: The transfer of thermal energy between objects
because their temperatures are different.
Heat: energy transfer
Symbol: Q
Units: Calorie (cal) or Joule (J)
1 cal = 4.186 J (energy needed to raise
1g of water by 10C)
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Heat transfer to an object
The amount of energy transfer Q to an object with mass m
when its temperature is raised by T:
Q = cmT
Change in
temperature
Energy transfer
(J or cal)
Mass of object
Specific heat
(J/(kgoC) or cal/(goC)
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Example
Substance
Specific Heat Specific Heat
J/kg oC
cal/g oC
aluminum
900
0.215
copper
387
0.092
water
4186
1.00
A 1 kg block of Copper is
raised in temperature by
10oC. What was the heat
transfer Q.?
Answer:
Q = cmT
= 387*1*10=3870 J
1 cal = 4.186 J
Q = 924.5 cal
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Another one
A block of Copper is dropped from a height of 10 m.
Assuming that all the potential energy is transferred into
internal energy (heat) when it hits the ground, what is the
raise in temperature of the block (ccopper=387 J/(kgoC))?
Potential energy: mgh = 10mg Joules
All transferred into heat Q:
Q
= cmT
10mg = 387mT
T = 10g/387 = 0.25oC
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Calorimetry
If we connect two objects with different temperature
energy will transferred from the hotter to the cooler
one until their temperatures are the same.
If the system is isolated:
Qcold=-Qhot
mcoldccold(Tfinal-Tcold)=-mhotchot(Tfinal-Thot)
the final temperature is: Tfinal=
mcoldccoldTcold+mhotchotThot
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mcoldccold+mhotchot
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Clicker Question!
A block of iron that has been heated to 1000C is dropped
in a glass of water at room temperature (200C).
After the temperatures in the block and the water
have become equal:
a) The water has changed more in temperature than the
iron block.
b) The water has changed less in temperature than the
iron block
c) the temperatures of both have changed equally
d) I need more info to say anything!
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Heating Water with a Ball of Lead
A ball of Lead at T=100oC with mass 300 g is dropped in a
glass of water (0.3 L) at T=200C. What is the final
(after thermal equilibrium has occurred) temperature of
the system? (cwater=1 cal/goC, clead=0.03 cal/goC water=103kg/m3)
Qcold=-Qhot
mwatercwater(Tfinal-Twater)=-mleadclead(Tfinal-Tlead)
mwatercwaterTwater+mleadcleadTlead
Tfinal=
mwatercwater+mleadclead
= (0.3*1*20+0.3*0.03*100)/(0.3*1+0.3*0.03)=
= 6.9/0.309=22.3oC
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An Example
The contents of a can of soda (0.33 kg) which
is cooled to 4oC is poured into a glass (0.1 kg) that is at
room temperature (200C). What will the temperature
of the filled glass be after it has reached full equilibrium
(glass and liquid have the same temperature)?
Given cwater=4186 J/(kgoC) and cglass=837 J/(kg0C)
Qcold=-Qhot
mwatercwater(Tfinal-Twater)=-mglasscglass(Tfinal-Tglass)
mwatercwaterTwater+mglasscglassTglass
Tfinal=
mwatercwater+mglasscglass
= (0.33*4186*4+0.1*837*20)/(0.33*4186+0.1*837)=
= 4.9oC
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And another
A block of unknown substance with a mass of 8 kg, initially
at T=280K is thermally connect to a block of copper (5 kg)
that is at T=320 K (ccopper=0.093 cal/g0C). After the system
has reached thermal equilibrium the temperature T equals
290K. What is the specific heat of the unknown material
in cal/goC?
Qcold=-Qhot
munknowncunknown(Tfinal-Tunknown)=-mcopperccopper(Tfinal-Tcopper)
cunknown=-mcopperccopper(Tfinal-Tcopper)
munknown (Tfinal-Tunknown)
cunkown=-5000·0.093·(290-320) =0.17 cal/goC
8000·(290-280)
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copper
????
13
Mixing 3 liquids
Three different liquids are mixed together in a calorimeter.
The masses, specific heats and initial temperatures of the liquids are:
T1 = 24.5 °C,
m1 = 475 g,
c1 = 225 J/kgC,
T2 = 53.5 °C,
m2 = 355 g,
c2 = 500 J/kgC,
T3 = 81.5 °C.
m3 = 795 g.
c3 = 840 J/kgC.
What will be the temperature of the mixture in Celsius?
Q1=-(Q2+Q3) [one could also use: Q2=-(Q1+Q3) or Q3=-(Q1+Q2)]
c1m1(Tf-T1)=- c2m2(Tf-T2)- c3m3(Tf-T3)
c1m1Tf+c2m2Tf+c3m3Tf= c1m1T1+c2m2T2+c3m3T3
c1m1T1+c2m2T2+c3m3T3
Tf=
=
c1m1+c2m2+c3m3
225x475x24.5+500x355x53.5+840x795x81.5
225x475+500x355+840x795
=69.9oC
Like with two substances, the final temperature is a weighted
average of T1,T2 and T3 with the c’s and m’s being the weights
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Phase Change
GAS(high T)
Q=cgasmT
Gas 
liquid
Q=csolidmT Solid (low T)
liquid (medium T)
liquid 
solid
Q=cliquidmT
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Gas 
liquid
Phase change
When heat is added to a liquid, potential energy goes to 0
(the energy stored in the stickiness of the liquid is taken away)
DURING THE CHANGE FROM LIQUID TO GAS, THE KINETIC
ENERGY DOES NOT CHANGE AND SO THE TEMPERATURE
DOES NOT CHANGE.
ALL ADDED HEAT GOES TO CHANGING PE
When heat is taken from a gas, potential energy goes to the
stickiness of the fluid
DURING THE CHANGE FROM GAS TO LIQUID, THE KINETIC
ENERGY DOES NOT CHANGE AND SO THE TEMPERATURE
DOES NOT CHANGE.
ALL REMOVED HEAT GOES TO CHANGING PE
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liquid
solid
Phase change
When heat is added to a solid to make a liquid, potential energy in
the bonds between the atoms become less
DURING THE CHANGE FROM SOLID TO LIQUID, THE KINETIC
ENERGY DOES NOT CHANGE AND SO THE TEMPERATURE
DOES NOT CHANGE.
ALL ADDED HEAT GOES TO CHANGING PE
When heat is taken from a liquid, the bonds between atoms
becomes stronger (potential energy is more negative)
DURING THE CHANGE FROM LIQUID TO SOLID, THE KINETIC
ENERGY DOES NOT CHANGE AND SO THE TEMPERATURE
DOES NOT CHANGE.
ALL REMOVED HEAT GOES TO CHANGING PE
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Clicker Question
Which of the following is not true?
a) During a phase transformation, the temperature of
a material remains unchanged
b) During a phase transformation, the kinetic energy of
the material remains unchanged
c) During a phase transformation, the potential energy
of the material is unchanged
d) During a phase transformation, the total energy of
the material changes
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Okay, the Temperature does not
change in a phase transition!
But what is the amount of heat added
to make the phase transition?
Gas 
liquid
Qgasliquid=-MLv
Qliquidgas=+MLv
M:mass
Lv=latent heat of vaporization (J/kg or cal/g)
depends on material.
Use the table in the book for LON-CAPA
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solid 
liquid
Qliquidsolid=-MLf
Qsolidliquid=+MLf
M:mass
Lf=latent heat of fusion (J/kg or cal/g)
depends on material.
Use the table in the book for LON-CAPA
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Phase Change
GAS(high T) Q=cgasmT
Gas 
liquid
Q=mLv
Q=csolidmT Solid (low T)
liquid (medium T)
Q=cliquidmT
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liquid 
solid
Q=mLf
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Clicker Question!
Ice is heated steadily and becomes liquid and then vapor.
During this process:
a) the temperature rises continuously.
b) when the ice turns into water, the temperature
drops for a brief moment.
c) the temperature is constant during the phase
transformations
d) the temperature cannot exceed 100oC
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ice
water
0
ice+water
steam
T (oC)
water+steam
100
THE PHASE TRANSFORMATIONS OF WATER
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steam
water+steam
water
ice+water
ice
Ice with T=-30oC is heated
to steam of T=1500C.
100
How much heat (in cal) has
T (oC)
been added in total?
cice=0.5 cal/goC
0
cwater=1.0 cal/goC
csteam=0.480 cal/goC
Lf=540 cal/g
Lv=79.7 cal/g
m=1 kg=1000g
A) Ice from -30 to 0oC
Q=1000*0.5*30= 15000 cal
B) Ice to water
Q=1000*540= 540000 cal
C) water from 0oC to 100oC
Q=1000*1.0*100=100000 cal
D) water to steam
Q=1000*79.7=
79700 cal
E) steam from 100oC to 1500C Q=1000*0.48*50=24000 cal
TOTAL
Q=
=758700 cal
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A Question
A block of gold (room temperature 200C)
is found to just melt completely after
supplying 4x103 J of heat. What
was the mass of the gold block?
Given: Lf=6.44x104 J/kg Tmelt=1063oC cspecific=129 J/kg0C
a)
b)
c)
d)
e)
0.01 kg
0.02 kg
0.03 kg
0.06 kg
10 kg
Q =cmT+mLf
= 129*m*1043+m*6.44x104
= 2x105*m
4000=2x105m
m = 0.02 kg
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How can heat be transferred?
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Conduction
Touching different materials: Some feel cold, others
feel warm, but all are at the same temperature…
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Thermal conductivity
metal
T=200C
wood
T=200C
The heat transfer
in the metal is
much faster than
in the wood:
(thermal
conductivity)
T=370C
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T=370C
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Heat transfer via conduction
Tc
Th
Conduction occurs if there is a
temperature difference between
two parts of a conducting medium
Rate of energy transfer P
A
P = Q/t (unit Watt)
P = kA(Th-Tc)/x = kAT/x
k: thermal conductivity
Unit:J/(msoC)
x
Metals
Gases
Nonmetals
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k~300 J/(msoC)
k~0.1 J/(msoC)
k~1
J/(msoC)
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Th
Tc
A
x
Example
A glass window (A=4m2,x=0.5cm)
separates a living room (T=200C)
from the outside (T=0oC). A) What
is the rate of heat transfer through
the window (kglass=0.84 J/(msoC))?
B) By what fraction does it change
if the surface becomes 2x smaller
and the temperature drops to -200C?
A) P=kAT/x=0.84*4*20/0.005=13440 Watt
B) Porig=kAT/x Pnew=k(0.5A)(2T)/x=Porig
The heat transfer is the same
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Another one.
An insulated gold wire (I.e. no heat lost to the air) is at
one end connected to a heat reservoir (T=1000C) and at the
other end connected to a heat sink (T=200C). If its length
is 1m and P=200W what is its cross section (A)?
kgold=314 J/(ms0C).
P=kAT/x=314*A*80/1=25120*A=200
A=8.0E-03 m2
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Water 0.5L
1000C
And another
A=0.03m2 thickness: 0.5cm.
1500C
A student working for his exam feels hungry and starts boiling
water (0.5L) for some noodles. He leaves the kitchen when
the water just boils.The stove’s temperature is 1500C.
The pan’s bottom has dimensions given above. Working hard
on the exam, he only comes back after half an hour. Is there
still water in the pan? (Lv=540 cal/g, kpan=1 cal/(ms0C)
To boil away 0.5L (=500g) of water: Q=Lv*500=270000 cal
Heat added by the stove: P=kAT/x=1*0.03*50/0.005=
=300 cal
P=Q/t t=Q/P=270000/300=900 s (15 minutes)
He’ll be hungry for a bit longer…
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Clicker Quiz!
Given your experience of what
feels colder when you walk on it,
a)
a rug
which of the surfaces would
b)
a steel surface
have the highest thermal
c)
a concrete floor
conductivity?
d)
has nothing to do with
thermal conductivity
The heat flow rate is k A (T1 − T2)/x. All things being
equal, bigger k leads to bigger heat loss.
From the book: Steel = 40, Concrete = 0.84,
Human tissue = 0.2, Wool = 0.04, in units of J/(s.m.C°).
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Multiple Layers
Th k
1
Tc
k2
A
L1
Q A(Th  Tc )
P=
=
t  ( Li / ki )
i
L2
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Q A(Th  Tc )
P=
=
t  ( Li / ki )
i
inside
Insulation
Th
Tc
L1 L2 L3
A house is built with 10cm thick wooden walls and roofs.
The owner decides to install insulation. After installation
the walls and roof are 4cm wood+2cm insulation+4cm wood.
If kwood=0.10 J/(ms0C) and kisolation=0.02 J/(ms0C), by what
factor does he reduce his heating bill?
Pbefore=AT/[0.10/0.10]=AT
Pafter=AT/[0.04/0.10+0.02/0.02+0.04/0.10]=0.55AT
Almost a factor of 2 (1.81)!
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Convection
T high
 low
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Radiation
Nearly all objects emit energy through radiation:
P=AeT4 : Stefan’s law (J/s)
=5.6696x10-8 W/m2K4
A: surface area
e: object dependent constant emissivity (0-1)
T: temperature (K)
P: energy radiated per second.
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emissivity
Ideal reflector
e=0
no energy is absorbed
Ideal absorber (black body)
e=1
all energy is absorbed
also ideal radiator!
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Black body
A black body is an object that absorbs all electromagnetic radiation
that falls onto it. They emit radiation, depending on their temperature.
If T<700 K, almost no visible light is produced (hence a ‘black’ body).
The energy emitted from a black body: P=T4 with =5.67x10-8 W/m2K4
b=2.8977685(51)×10−3 mK
Wiens displacement constant
More in PHY232!
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Infrared Radiation
The human body emits radiation in the infrared.
With a body temperature of T=(273+37 K) = 310 K
the wavelength of the thermal emission is ~5E-5 m
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A BBQ
The coals in a BBQ cover an area of 0.25m2. If the
emissivity of the burning coal is 0.95 and their
temperature 5000C, how much energy is radiated every
minute?
P = AeT4 J/s
= 5.67x10-8*0.25*0.95*(773)4=4808 J/s
1 minute: 2.9x105 J (to cook 1 L of water 3.3x105 J)
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Net Power Radiated
If an object would only emit radiation it would eventually
have 0 K temperature. In reality, an object emits AND
receives radiation.
PNET = Ae (T4-T04)
where
T: temperature of object
T0: temperature of surroundings.
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Example
The temperature of the human body is 370C. If the
room temperature is 200C, how much heat is given
off by the human body to the room in one minute?
Assume that the emissivity of the human body is 0.9
and the surface area is 2 m2.
P=Ae(T4-T04)
=5.67x10-8 * 2 * 0.9 *(310.54-293.54)=
=185 J/s
Q=P*T=185*60=1.1x104 J
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For Next Week
Chapter 14:
The Laws of Thermodynamics
Homework Set 11 Due 11/28
Covers Chapter 13
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