heat 14 (01 of 32)
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Heat 14 (01 of 32)
Physics Lecture Notes
Heat
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
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 is random thermal Energy
Heat 14 (13 of 32)
Heat is a form of energy and can be equated to mechanical energy.
Heat As Energy Transfer
cal 1J 186.4
calk 1J 4186
The apparatus below is used to determine the mechanical equivalent of heat:
Heat 14 (13 of 32)
Heat As Energy Transfer
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
Definition of heat:
Heat is thermal energy transferred from one object to another because of a difference in temperature.
Heat 14 (13 of 32)
m
TmcQ
(Q) Thermal energy added
(m) Mass of the object
(c) Specific heatof the material
(T) Change intemperature
Heat As Energy Transfer
Heat 14 (13 of 32)
Chapter 14Page 387
Heat 14 (13 of 32)
A 50 g piece of cadmium is at 20 oC. If 400 J of heatis added to the cadmium, what is its final temperature
TmcQ
mcQ
T
oCkg
J 230 kg 050.0
J 400
oC 8.34
TTT of
oof C 8.34C 0.20T C 8.54 o
Heat As Energy Transfer
ocadmium Ckg
J 230c
Problem:
Heat 14 (13 of 32)
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?
2
KΔQ bullet
bullet
4mv
Tmc2
c4v
T2
o
2
Ckg
J 031 4
m/s 300
oC 173
Heat As Energy Transfer
olead Ckg
J 130c
Problem:
Heat 14 (13 of 32)
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?
blockblock QU
Tmcmgh
gTc
h
2
oo
m/s 8.9
C50.0 Ckg
J 450
m 23
Heat As Energy Transfer
oiron Ckg
J 450c
Problem:
Heat 14 (13 of 32)
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?
blockblock K85.0Q
2
mv85.0Tmc
2
c2
v85.0T
2
o
2
Ckg
J 390 2
m/s 3085.0
C 0.1 o
Heat As Energy Transfer
ocopper Ckg
J 390c
Problem:
Heat 14 (13 of 32)
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.
lossgain QQ
IIIWWW TΔcmTΔcm
ff T5004504.022T418620
ff T180900001841840T83720
C 23T of
CalorimetryProblem:
owater
oiron
Ckg
J 4186c
Ckg
J 450c
Heat 14 (13 of 32)
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
lossgain QQ
CCCGGGWWW TΔcmTΔcmTΔcm
fff T7870206804T25245208T4.1674
C 5.29T of
Calorimetry
fff T903902.027T8403.027T41864.0
Problem:
oglass
owater
ocopper
Ckg
J 840c
Ckg
J 4186c
Ckg
J 390c
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 problem1. Draw sketch of problem.
2. List given 2. List given information.information.Mug massMug mass mmmm = = 0.500 kg0.500 kg
Coffee massCoffee mass mmcc = = 0.200 kg0.200 kg
Initial temperature of coffee and mug:Initial temperature of coffee and mug: tt00 = 20 = 2000CCFinal temperature of coffee and mug:Final temperature of coffee and mug: ttff = 96 = 9600CC
Total heat to raise temperature
of coffee (water) and mug to 960C.
3. List what is to be 3. List what is to be found: found:
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:4. Recall applicable formula or law:Q = mc tHeat Gain or Loss:
5. Decide that TOTAL heat is that 5. Decide that TOTAL heat is that required to raise temperature of required to raise temperature of mug and water (coffee). Write mug and water (coffee). Write equation.equation.
QQTT = = mmmmccmm t + mt + mwwccw w tt
6. Look up specific 6. Look up specific heats in tables:heats in tables:
Copper: cCopper: cmm = 390 J/kg C = 390 J/kg C00 Coffee (water): cCoffee (water): cww = 4186 J/kg = 4186 J/kg
CC00
t = 960C - 200C = 76 C0
t = 960C - 200C = 76 C0
Water: (0.20 kg)(4186 J/kgC0)(76 C0)
Cup: (0.50 kg)(390 J/kgC0)(76 C0)
QT = 63,600 J + 14,800 J QT = 78.4 kJ
QT = 78.4 kJ
7. Substitute info / solve problem:7. Substitute info / solve problem:
QT = mmcm t + mwcw t
Copper: cCopper: cmm = 390 J/kg C = 390 J/kg C00 Coffee (water): cCoffee (water): cww = 4186 J/kg C = 4186 J/kg C00
Example 1(Cont.): How much heat needed to heat cup and coffee from 20 to 960C?
mc = 0.2 kg; mw = 0.5 kg.
m1 m2
TH TL
lossQ gainQ
Conservation of thermal energy:
lossgain QQ
111222 TΔcmTΔcm fH11Lf22 TTcmTTcm
Final Temperature:
2211
L22H11f cmcm
TcmTcmT
Calorimetry
Heat 14 (13 of 32)
mLQ
(Q) Thermal energy added
(L) Latent heatof the fusion or
vaporization
(m) Mass of the object
Latent Heat - Stored / Hidden
Energy is required for a material to change phase,
Even though its temperature is not changing.
Heat 14 (13 of 32)
The water problem
SubstanceLatent Heat
FusionkJ/kg
MeltingPoint
°C
Latent HeatVaporization
kJ/kg
BoilingPoint
°C
Alcohol, ethyl 108 −114 855 78.3
Ammonia 339 −75 1369 −33.34
Carbon dioxide 184 −78 574 −57
Helium 21 −268.93
Hydrogen(2) 58 −259 455 −253
Lead[8] 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 293
Water 334 0 2260 100
Table of latent heatsThe following table shows the latent heats and change of phase temperatures of some common fluids and gases.
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 vaporChapter 14 - Page 392
Heat 14 (13 of 32)
50
Q2= mLf
3.33 x 105 J
Q1= mcIt
1.05 x 105 J
Q4= mLv
22.6 x 105 J
Q5= mcSt
1.01 x 105 J
Q3= mcWt
4.19 x 105 J
Latent Heat
1 kgIce
-50 oC
Heat 14 (13 of 32)
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
Heat required to convert 1 kg of ice at -50 oC to steam at 150 oC
3.22 x 106 J
Latent Heat
Heat 14 (13 of 32)
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?
loss)gain( QQ
AAAfI TcmLm
f
AAAI L
Tcmm
kgJ
10 x 33.3
C300Ckg
J900 kg 4.0
m5
oo
I
kg 032.0
Latent Heat
Problem:
owaterf
oaluminum
Ckg
kJ 333L
Ckg
J 900c
Heat 14 (13 of 32)
Conduction
Heat conduction can be visualized as occurring through molecular collisions.
The heat flow per unit time is given by:
x
x
TTkA
tQ 21
Heat 14 (13 of 32)
The constant k is called the thermal conductivity.
Materials with large k are called conductors; those with small k are called insulators.
Conduction
Chapter 14Page 396
Heat 14 (13 of 32)
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.
xTkA
tQ
m 10 x 0.3
Co 202
cm 100m 12cm 310 x 6.1
Com
W 84.0
3
s/J 896tΔQΔ
Conduction
Problem:
oglass Cms
J 84.0k
Heat 14 (13 of 32)
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?
xTkA
tQ
x
tTkAQ
m .0060
s 3600oC 25 2m 3 oCm
W 84.0
J 710 x 78.3QΔ
Conduction
oglass Cms
J 84.0k
Heat 14 (13 of 32)
Building materials are measured using R−values rather than thermal conductivity:
Where, L is the thickness of the material.
Conduction
kL
R
Chapter 14Page 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,
nRTNkTU23
23
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.
Specific heat of a substance is the energy required to change the temperature of a fixed amount of matter by 1° C.
TΔmcQ
Heat 14 (13 of 32)
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.
mLQ
Heat 14 (13 of 32)
Summary
Heat transfer takes place by conduction, convection, and radiation.
In conduction, energy is transferred through the collisions of molecules in the substance.
xTkA
tQ
Heat 14 (13 of 32)
Summary
Internal Energy
Internal energy of an ideal (monatomic) gas: 2
21 vm NU
NkTU23
nRTNkT nRTU
23
kTvm232
21
kinetic energy in terms of the temperature
Heat 14 (13 of 32)
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