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1. CHAPTER V
TEMPERATURE AND HEAT

2. A. TEMPERATURE
The tool used to measure the temperature a thermometer called there are
three kind of thermometer namely : Celsius, reamer, Fahrenheit.
100 Scale
80 Scale
180 Scale
1000
320
800
R
2120
00 00
C
Melting point
F
C = thermometer Celsius (boiling point = 1000
C )
R = thermometer Reamer (boiling point = 800
R )
F= thermometer Fahrenheit (boiling point = 2120
F )
K= thermometer Kelvin (boiling point= 373 K)
Boiling point
K
273
373
100 Scale

3. The Formula :
tC : tR : ( tF – 32 ): (tK – 273) = 100 : 80 : 180: 100
tC : tR : ( tF – 32 ) : (tK - 273) = 5 : 4 : 9 : 5
The unit of absolute temperature is Kelvin.
T = tC + 273
The general formula:
mYbY
mY
mXbX
mX
tt
ttY
tt
ttX
−
−
=
−
−
tX = temperatur of thermometer X
tmX= melting point of thermometer X
tbX = boiling point of thermometer X
tY = temperatur of thermometer Y
tmY= melting point of thermometer Y
tbY = boiling point of thermometer Y

4. Problems.
1. Scale of the Celsius thermometer show 500
.
What is the scale of the thermometer :
a. Kelvin
b. Reamur
c. Fahrenheit
2. a. 73 K = ….o
F b. 40 o
R= …O
K c. 45O
F=…o
K
3. If X scale based on Celsius scale and have melting point 100
and
boiling point 1400
. determine Celcius scale if X scale 400
.
4. Thermometer B have a melting point – 50
and boiling point 1150
.
determine the temperature of B thermometer, if the Fahrenheit
scale is 1130
F.
5. Find the Celsius scale which is equal to the Fahrenheit scale.
Find also the reamur scale which is equal to the Fahrenheit scale.
6. Find the Fahrenheit scale which is bigger 200
than the reamer
scale. Find also the Celsius scale which is smaller 100
than the
Fahrenheit.

5. B. HEAT
Is the energy transferred from one object to the other of the temperature
differences.
The amount of energy absorbed or released of an object can be state with
the formula:
Q = m . c . t∆
C = m . c
Q = C . t∆
Q = heat (J)
m = mass (kg)
c = specific heat of substance (J/kg K)
C = heat capacity (J/kg)
t = temperature changes (K)∆
1 cal = 4,2 J
1 cal/g.o
C = 4200 J/kg K

6. Problems :
1. An amount of 3.106
j heat is given to the 4 kg aluminum box with the
temperature of 300
C ( c = 900 J / kg 0
c ). Find the end temperature of the
box.
2. An object of 400 g with a specific heat of 3600 J / kg0
c is heated so the
temperature raise from 250
C to 1000
C.
Find : a. Heat capacity.
b. the absorb heat.
3. An aluminum foil of 200 g with a temperature of 900
C is put inside 100 g
water of 200
C. Suppose that there is no missing heat. cal = 900 J / kg 0
C, cw
= 4200 J / kg 0
C find the end temperature of the mixture.
4. A 0,5 kg tin of 1000C plunge into 0,2 kg water of 200
C in an aluminum
calorie meter of 0,1 kg. The end of the temperature is 240
C.
cAl = 900 J / kg k, ca = 4200 J / kg k.
find the tin specific heat ?
5. Ice of 50 g at -10o
C is put on a 100 g hot plate of glass at 300o
C, find the
final temperature of the system. (cice= 0,5 cal/g o
C, Lice= 80 cal/g, cwater= 1 cal/g
o
C, and cglass= 0,2 cal/g o
C)

7. C. PHASE CHANGES OF SUBSTANCES
Graphic :
Gas
Solid
(padat)
Liquid
(Cair)
Q = m . L L = Latent Heat (J/kg)

8. Q1
00
Q2 Q3
1000
C
Q4
1000
C
00
 Solid absorb heat energy of Q1 = m . cs.Δt
 Melting absorb heat energy of Q2 = m . Lm
 Liquid absorb heat energy of Q3 = m . cl . Δt
 Boiling absorb heat energy of Q4 = m . Lb
 gas absorb heat energy of Q5 = m. cg . Δt
notice: Lm = Melting latent heat
L = Boiling latent heat
The changes of states especially for water from solid ( ice ) to gas
( steam ) caused by giving heat energy can explain as in the graph
below :

9. The problem of heat is on the black’s principle which is state that the
amount of the absorb heat is the same with the release heat.
Heat lost by hot object = Heat gained by cold object
In symbols,
C. BLACK’S PRINCIPLE
Q release = Q absorb

10. Problems
1. Find the heat energy to change 4 Kg of ice – 100
C to gas of 1000
C
ci = 2100 J/Kg k Le = 2256.103
J / kg
cw= 4200 J / kg k Lf = 334.103
J / kg
2. 25 gr of solid matter absorb heat energy of 200 J/s so it change all to gas
in the graph :
0
20 60 75 90
A
time ( minute )
Find specific latent heat of fusion and evaporation!
B C
D E
F

11. 3. 400 g ice of -40
C put into 800 g water of 600
C so the ice all
melted. Find the final temperature of the mixture.
4. A glass tube (m=300g) of specific heat 0,2 cal / g0
C filled with ice
of 25 g -20
C. Then water of 150 g 400
C put into the tube so that all
the ice are melted.
Find the final temperature of the mixture.

12. D. Substance Expansion
Any substance which increase the temperature will be expanded,
exceptions on water from 0°C to 4°C which will we reduce in volumes. This
stronger feature is called water anomaly.
I. Solid Expansion
Linear Expansion
Any solids with elongation ( eg.wire ) when it heated the length will be in
creased.
The elongations of the expansion can be found wit the formulate :
= Length elongation
= linier expansion coefficient
= Temperature
= Initial length
= Final length
∆
α
t∆
ο
t
t)(1ot
t.
∆+=
∆=∆
α
α

 o

13. Problems
1. Length of a railway is made of steel is 20 m in 100
C. Find the
length of the railway when the temperature is 400
C.(α = 11.10-6
/C)
2. Two railway, each of then 15 m length will be set. If the minimum
and maximum temperature on that place is 250
C and 400
C,
find the minimum distance two railway to be set. (α = 11.10-6
/C
3. A bar of steel (l = 2m) increase the length 2 mm when heated
arrive the temperature increase 600
C. Find the length expansion
of the steel when the temperature increase 200
C.

14. II. Area Expansion
A plate when heated will be increase in area. The area
expansion can be found by the formula :
= Area expansion
= Area expansion coeff
= Initial area
= final area
( )
αβ
βο
οβ
2
1
=
∆+Α=Α
∆Α=∆Α
tt
t ∆Α
β
οΑ
tΑ

15. Problems
1. A plaited of aluminum with its length of 20 cm and wide
12 cm heated from 200
C to 500
C.
Find the final area of the plate.(α= 24.10-6
/C)
2. A wheel of steel of 200
C have an inside radius of 200
mm will be set to a wooden wheel of radius 202 mm it
the linear expansion coefficient the steel 1,25.10-5
/0
C.
Find the final temperature so that the wheel of steel
will be set up to the wooden wheel precisely.

16. III. Liquids Expansion
Liquids when heated will increase in
volum,exceptionnally of water. When water heated
from 00
C to 40
C the volume will reduce.
The liquids expansion can be found from the
formula :
= Volume expansion
= Volume expansion coefficient
= Final volume
= Initial volume
( )
∝=
∆+=
∆=∆
3
1VV
VV
γ
γ
γ
tt
t
ο
ο
V∆
γ
tV
oV

17. Problem:
1. The bar of steel of 20 cm x 5 cm x 3 cm temperature
400
C. The bar is heated so it’s final volume 300,54 cm3
. If
the linear expansion coefficient of steel 15.10-6
/0
C. Find
the final temperature the bar heated !
2. A bottle have a volume of 400 cm3
filled with water of
400
C. Then the bottle heated to 600
C. Find the water
spilled out if the bottle expansion: (α glass= 9.10-6
/o
C, γ
water= 2,1.10-4
/o
C)
a. Neglected
b. Counted,

18. IV. Gas Expansion
When a gas in a closed room the temperature remain
constant, according to Boyle, the changes of pressure
and volumes is always constant.
According to Gay-Lussac, if the pressure is constant we
have the relationship as follow :
constanP.V =
konstan
T
V
=

19. If the formula (1) and (2) combine, the
formula is know as Boyle-Gay Lussac
formula.
= First Pressure ( atm,N
/m2
)
= Last Pressure ( atm,N
/m2
)
= First Volume ( cm3
,m3
)
= Last Volume ( cm3
,m3
)
= First Temperature
= Last Temperature
1atm = 76 cm Hg
konstan
T
PV
=
2
22
1
11
T
VP
T
VP
=⇒
1P
1V
2V
1T
2T
2P

20. Problems
1. A gas of H2 in a tube of 270
C and volume 6 L, have the
pressure of 2 atm. Find the volume if the gas heated to
1770
C in pressure constant.
2. A tube with it’s volume 8 L filled with gas of O22. The
gas is heated to constant temperature so the pressure
change to 3 times as before. Find the percentage of
the volume changes ?.

21. E. Heat Transfer.
Heat can travel in three ways :
1. Conduction
Conduction is the flow of heat through matter without the changes of the particles of
the medium.
eg : heat transfer on a bar of iron.
The amount of the heat energy conduction per second can be found in the formula
Q = Heat energy
t = Time
K = thermal conductivity
A = Area
d = Length of the rod
T = Temperature Changes
Two rods of different type join together will follow the rules that the heat energy transfer per
second on the two rods have the same amount as in the formula below :
d
TAK
t
Q ∆
=
..
∆
Q1 Q2
=
t t

22. 2. Convection
Convection is the flow at heat with the changes of the particles of the
medium.
eg : heat transfer of water heating.
The amount of the heat energy per second is :
3. Radiation
Radiation is the flow of heat in the form of electromagnetic waves ( no
medium melded ). The amount of energy radioactive is determined kg
Stefan Bolliztman :
Q = Radiation energy A = Area of surface
t = time T = Temperature
σ = 5,67.10-8
watt / m2
k4
e = Emisivity
TAh
t
Q
∆= ..
h = Convection coefficient
4
ΑΤ= σe
t
Q

23. Black body is an object that are able to amity / absorb of heat energy perfectly.
If the temperature of the body is different we than the surroundings, if follows :
Problems
1. An air conditional room have a glass window of 4 m2
area end the width of 2 mm. if
the temperature of the surface inside is 200
C end outside 300
C, find the heat
energy per second.( kglass= 0,8 J / m.s.o
C )
Two rods of P and Q jointed. If Kp = ½ KQ and AC = 2 CB, find the temperature of
C.
3. An object of 2 m2
300
C put in on a room of 200
C find the heat energy emit from the
object through convection in 5 minutes.
).(
4
2
4
1 Τ−ΤΑ= σe
t
Q
P Q
A
400
C
C B
1100
C
2.
h = Convection coefficient =8 J/s.m2
.o
C

24. 4. A string of wire on a filament lamp has an area of 100 mm2
11270
C. It the wire supposed to be a black body find :
a. The heat energy radiate each second.
b. the electric current flow if the lamp connected to 220 V.
5. An object has an area 10 cm2
radiates on 5270
C in a room of
1270
C. The object emisivity of 0,8.
Find then energy radiates in 1 minute heat.