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Transcript
```Heat
Physics Lecture
Notes
Heat 14 (01 of 32)
Heat
1) Heat As Energy Transfer
2) Internal Energy
3) Specific Heat
4) Calorimetry
5) Latent Heat
6) Heat Transfer: Conduction
Heat 14 (02 of 32)
Heat As Energy Transfer
Heat is random thermal Energy
Unit of heat: calorie (cal)
1 cal is the amount of heat necessary to raise the
temperature of 1 g of water by 1 Celsius degree.
1 kcal is the amount of heat necessary to raise the
temperature of 1 kg of water by 1 Celsius degree.
Heat 14 (13 of 32)
Heat As Energy Transfer
Heat is a form of energy and can be equated to
mechanical energy.
The apparatus below is used to determine the
mechanical equivalent of heat:
4 .186 J  1 cal
4186 J  1 kcal
Heat 14 (13 of 32)
Heat As Energy Transfer
Definition of heat:
Heat is thermal energy transferred from one
object to another because of a difference in
temperature.
The sum total of all the energy of all the molecules in a
substance is its internal (or thermal) energy.
Temperature: measures molecules’ average kinetic energy
Internal energy: total energy of all molecules
Heat: transfer of energy due to difference in temperature
Heat 14 (13 of 32)
Heat As Energy Transfer
(c) Specific heat
of the material
(Q) Thermal
Q  mc T
(m) Mass of
the object
m
(T) Change in
temperature
Heat 14 (13 of 32)
Chapter 14
Page 387
Heat 14 (13 of 32)
Heat As Energy Transfer
Problem:
A 50 g piece of cadmium is at 20 oC. If 400 J of heat
J
kg  Co
Q  mc T
Q
T 
mc

400 J
0.050 kg  230
J
kg  Co
 34.8 C o
Tf  To   T
Tf  20 .0 o C  34 .8 C o
 54.8 o C
Heat 14 (13 of 32)
Heat As Energy Transfer
Problem:
A 100 g lead bullet traveling at 300 m/s is stopped by a large
tree. Half the kinetic energy of the bullet is transformed into
heat energy and remains with the the bullet while the other
half is transmitted to the tree. What is the increase in
temperature of the bullet?
J
o
kg

C
ΔK bullet 
Q bullet   
2
mv 2
mcT 
4
2
v
T 
4c

300 m/s 2

4  130 J o
 173 C o
kg  C
Heat 14 (13 of 32)
Heat As Energy Transfer
Problem:
A 3.0 kg block of iron is dropped from rest from the top
of a cliff. When the block hits the ground it is observed
that its temperature increases by 0.50 oC. Assume that
all the potential energy is used to heat the block. How
high is the cliff?
J
c iron  450
kg  Co
  U block    Q block 
mgh  mc T
cT
h
g

J
450
0.50o C
kg  Co

9.8 m/s 2

 23 m
Heat 14 (13 of 32)
Heat As Energy Transfer
Problem:
A 1.5 kg copper block is given an initial speed of 30 m/s on
a rough horizontal surface Because of friction, the block
finally comes to rest. If the block absorbs 85% of its initial
kinetic energy in the form of heat, Calculate its increase in
temperature?
J
ccopper  390
o
kg

C
 Q block    0.85  K block 
mv 2
mcT  0.85 
2
v2
T  0.85 
2c

30 m/s 2
 0.85 
2  390 J
 1 .0 o C
kg  Co
Heat 14 (13 of 32)
Calorimetry
Problem:
A 0.40 kg iron horseshoe that is initially at 500 oC is
dropped into a bucket containing 20 kg of water at 22 oC.
What is the final equilibrium temperature?
Neglect any heat transfer to for from the surroundings.
Q gain    Q loss 
m W c W Δ TW   m I c I Δ TI
20 4186 Tf
J
c iron  450
kg  Co
J
c water  4186
kg  Co
 22   0 .4 450 500  Tf 
83720Tf  1841840  90000  180 Tf
Tf  23 o C
Heat 14 (13 of 32)
Calorimetry
Problem:
A 200 g block of copper at a temperature
of 90 oC is dropped into 400 g of water at
27 oC. The water is contained in a 300 g
glass container. What is the final
temperature of the mixture
J
kg  Co
J
c water  4186
kg  Co
J
c glass  840
kg  Co
ccopper  390
Q gain    Q loss 
m W c W Δ TW  m G c G Δ TG   m C c C Δ TC
0.4 4186 Tf
 27   0 .3 840 Tf  27   0 .2 390 90  Tf 
1674 .4 Tf  45208  252 Tf  6804  7020  78 Tf
Tf  29 .5 o C
Heat 14 (13 of 32)
Example 1: A 500-g copper coffee mug
is filled with 200-g of coffee. How much
heat was required to heat cup and
coffee from 20 to 960C?
1. Draw sketch of problem.
2. List given
information.
Mug mass mm = 0.500 kg
Coffee mass mc = 0.200 kg
Initial temperature of coffee and mug: t0 = 200C
Final temperature of coffee and mug: tf = 960C
3. List what is to be Total heat to raise
temperature
found:
of coffee (water) and mug to 960C.
Example 1(Cont.): How much heat needed to heat
cup and coffee from 20 to 960C?
mm = 0.2 kg; mw = 0.5 kg.
4. Recall applicable formula or law:
Heat Gain or Loss: Q = mc t
5. Decide that TOTAL heat is that
required to raise temperature of
mug and water (coffee). Write
equation.
QT = mmcm t + mwcw t
6. Look up specific Copper: cm = 390 J/kg C0
heats in tables: Coffee (water): cw = 4186 J/kg
C0
Example 1(Cont.): How much heat needed
to heat cup and coffee from 20 to 960C?
mc = 0.2 kg; mw = 0.5 kg.
7. Substitute info / solve problem:
Copper: cm = 390 J/kg C0
Coffee (water): cw = 4186 J/kg C0
QT = mmcm t + mwcw t
Water: (0.20 kg)(4186 J/kgC0)(76
t = 960C - 200C
0
C)
0
=
76
C
Cup: (0.50 kg)(390 J/kgC0)(76 C0)
QT = 78.4
QT = 63,600 J + 14,800 J
kJ
Calorimetry
m1
TH
Q loss 
m2

TL
Q gain 
Conservation of thermal energy:
Q gain    Q loss 
m 2 c 2 Δ T2   m 1c1Δ T1
m 2 c 2 Tf  TL   m 1c1 TH  Tf 
Final Temperature:
m1c1TH  m 2c 2 TL
Tf 
m1c1  m 2c 2
Heat 14 (13 of 32)
Latent Heat - Stored / Hidden
Energy is required for a
material to change phase,
Even though its temperature
is not changing.
(m) Mass of
the object
Q  mL
(Q) Thermal
(L) Latent heat
of the fusion or
vaporization
Heat 14 (13 of 32)
The water problem
Table of latent heats
The following table shows the latent heats and change of phase temperatures of some common fluids and gases.
Substance
Latent Heat
Fusion
kJ/kg
Melting
Point
°C
Latent Heat
Vaporization
kJ/kg
Boiling
Point
°C
Alcohol, ethyl
108
−114
855
78.3
Ammonia
339
−75
1369
−33.34
Carbon dioxide
184
−78
574
−57
21
−268.93
Helium
Hydrogen(2)
58
−259
455
−253
24.5
327.5
871
1750
Nitrogen
25.7
−210
200
−196
Oxygen
13.9
−219
213
−183
R134a
−101
215.9
−26.6
Toluene
−93
351
110.6
Turpentine
Water
293
334
0
2260
100
Latent Heat
Heat of fusion, Lf: heat required to change 1.0 kg
of material from solid to liquid
Heat of vaporization, Lv: heat required to change 1.0 kg
of material from liquid to vapor
Chapter 14 - Page 392
Heat 14 (13 of 32)
Latent Heat
1 kg
Ice
-50 oC
Q2= mLf
Q4= mLv
3.33 x 105 J
22.6 x 105 J
150
100
50
0
50
Q1= mcIt
1.05 x 105 J
Q3= mcWt
Q5= mcSt
4.19 x 105 J
1.01 x 105 J
Heat 14 (13 of 32)
Latent Heat
Heat required to convert 1 kg of ice
at -50 oC to steam at 150 oC
Q1 = 1.05 x 105 J
Q2 = 3.33 x 105 J
Q3 = 4.19 x 105 J
Q4 = 22.6 x 105 J
Q5 = 1.01 x 105 J
3.22 x 106 J
Heat 14 (13 of 32)
Latent Heat
Problem:
A large block of ice at 0 oC has a hole chipped in it, and
400 g of aluminum pellets at a temperature of 30 oC are
poured into the hole. How much of the ice melts?
Q ( gain )   Q loss 
m I L f   m A c A  TA
J
caluminum  900
kg  Co
kJ
L f water   333
kg  Co
m A c A TA
mI  
Lf
J
0.4 kg  900

0  30 Co
kg  Co
mI  
J
5
3.33 x 10
kg
 0.032 kg
Heat 14 (13 of 32)
Conduction
Heat conduction can be visualized as occurring through
molecular collisions.
x
The heat flow per unit time is given by:
Q kAT1  T2 

t
x
Heat 14 (13 of 32)
Conduction
The constant k is called the thermal
conductivity.
Materials with large k are called
conductors; those with small k are
called insulators.
Chapter 14
Page 396
Heat 14 (13 of 32)
Conduction
Problem:
A window has a glass surface of 1.6 x 103 cm2 and a
thickness of 3.0 mm. Find the rate of heat transfer by
conduction through this pane when the temperature of
the inside surface of the glass is 20 oC and the outside
temperature is 40 oC.
J
k glass  0.84
s  m  Co
Q kAT

t
x
2
W 
1
m



 0.84
 1.6 x 10 3 cm 2 
 20 oC
oC 
 100 cm 

m

3.0 x 10  3 m




ΔQ
 896 J / s
Δt
Heat 14 (13 of 32)
Conduction
Problem:
A glass window pane has an area of 3.0 m2 and a
thickness of 0.60 cm. If the temperature difference
between its faces is 25 oC, how much heat flows
through the window per hour?
J
k glass  0.84
o
s

m

C
Q kAT

t
x
W 

 0.84
 3 m 2 25 Co 3600 s 

kAT t  
m  Co 
Q 
0.006 m
x



ΔQ  3.78 x 10 7 J
Heat 14 (13 of 32)
Conduction
Building materials are measured using R−values rather than
thermal conductivity:
L
R
k
Where, L is the thickness of the material.
Chapter 14
Page 397
Heat 14 (13 of 32)
Summary
Internal energy U refers to the total energy of all
molecules in an object. For an ideal monatomic gas,
U  3 NkT  3 nRT
2
2
Heat is the transfer of energy from one object to
another due to a temperature difference. Heat can be
measured in joules or in calories.
Q  mcΔ T
Specific heat of a substance is the energy required to
change the temperature of a fixed amount of matter
by 1° C.
Heat 14 (13 of 32)
Summary
In an isolated system, heat gained by one part of
the system must be lost by another.
Calorimetry measures heat exchange quantitatively.
Energy in involved in phase changes even though
the temperature does not change.
Heat of fusion: amount of energy required to
melt 1 kg of material.
Heat of vaporization: amount of energy required to
change 1 kg of material from liquid to vapor.
Q  mL
Heat 14 (13 of 32)
Summary
Heat transfer takes place by conduction,
In conduction, energy is transferred through the
collisions of molecules in the substance.
Q kAT

t
x
Heat 14 (13 of 32)
Internal Energy
Internal energy of an ideal
(monatomic) gas:
2
U  N 1 mv2

kinetic energy in terms
of the temperature
1 mv2
2
 3 kT
2
NkT  nRT
U  3 NkT
2
U  3 nRT
2
Heat 14 (13 of 32)
```
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