# PHYSICS CLASS: HEAT ENERGY

Heat is defined as a form of energy which flows due to
temperature difference. Temperature is defined as the degree of hotness or
coldness of a body. It is measured with an instrument known as thermometer.

**EFFECT OF HEAT**

Change
in temperature

Change
in state

Expansion

Emission
of electrons

Resistance
change

**EXPANSION**

Solids expands when heated and contracts when cooled

**ADVANTAGES OF EXPANSION**

It
is used to construct fire alarm

It
is used to construct bimetalic strip

It
can be used to remove a cork or stopper from a bottle without breaking the
stopper.

**LINEAR EXPANSIVITY**

The linear expansivity of a solid is defined as the increase
in length, per unit length for one degree rise in temperature.

Linear expansivity = increase in
length

Original
length x rise in temperature

Alpha (Î±) = L

_{2}-L_{1}
L1
x (Ó©

_{2}- Ó©_{1})
Where

Î± – co-efficient of linear expansively the 5.1 unit is (k

^{-1}) per Kelvin Initre
L

_{1}= length of the metal
L

_{2}= final length of the metal
Q

_{1}= initial temperature
Q

_{2}= fuel temperature**NOTE:**C

_{2 }– L

_{1}= Î”L

Q

_{2}– Q_{1}= Î”Î¸
Simplified 30 the linear expansively of a metal is o.000011
park what does this statement mean?

Solution

It means that a unit substance of the metal will expand by
0.000011 when it’s temperature changes by 1

^{0}K
Simplified 30 a telephone steel wire is 90m long at 0

^{0}K. how much longer will it be at 55^{0}k (linear expansively of steel is 0.000011k^{-1})
Solution

L

_{1}= 90m, or = 00k, L_{2}? Î˜2 = 55^{0}k
Î± = L

_{2}-L_{1}make L_{2}subject
L

_{1}Î¸_{2}-Î¸_{1}formula
L

_{2}= Î± L_{1}(Î¸_{2}-Î¸_{1}) + L_{1}
L

_{2}= 0.000011 x 90 (55-0) x 90
L

_{2}= 0.000011 x 4950 x 90
= 90.1m

Simplified 30 steel
bars of length 3m at 280k are to be used for constructing a reline. If the
linear expansively of steel is 1-0x10

^{-5}k^{-1}. What is the safety gap that must be left between successive bars if the highest temperature expected is 40^{0}k
Solution

L

_{1}= 3m, safety gap = Î”2 = ? Î¸_{1}= 28^{0}k
Î¸

_{2}= 400k, Î± = 1.0x10^{-3}k^{-1}
Î± = Î”L

L

_{1}Î”Î¸
Î”L =Î± L

_{1}Î”Î¸
= > 1.0x10

^{-5}x 3x(40-28)
= 3.6x10

^{-4}m
Simplified 31 A metal rod of length 40m at 200k heated to a
temperature of 450k if the new length of the rod is 40.05m calculate its linear
expansively

Solution

L

_{1}=40m, L_{2}= 40 05, Î¸_{1}= 20^{0}k,Î¸_{2}=45^{0}k
Î± =?

Î± = L2 – L1 = 40-05-40

L

_{1}Î¸_{2}-Î¸_{1}40x45-20
= 0.005 = 5x10

^{-5}k^{-1}
1000

Simplified 32 an iron rod is 100m at 10

^{0}C what must be the length of an aluminum rod at 0^{0}c, if the difference between length of the two rods is to remain the same at the same temperature (Î± fe = 12x10^{-6}, Î±Al = 24x10^{-6})
Solution

For iron

Î± fe = Î”Lfe

L

_{1}Î”Î¸fe
Î”Lfe =Î±feL

_{1}Î”Î¸fe
For aluminum

Î±AL = Î”LAL

L

_{1}Î”Î¸AL
Î”LAL =Î± AL L

_{1}AÎ¸AL
But Î”Lfe = Î”LAL

Î± fe L

_{1}Î”Î¸ = Î±ALL_{1}Î”Î¸
Î±feL

_{1}= Î± ALL_{1}AL
life = 100m

Î±fe = 12x10

^{-6}
L

_{1}AL = ?
Î±Al = 24x10

^{-6}
= 12 x10

^{-6}x 100 = 24x10^{-6}x LINL
12x10-6 x 10 = LIAL

24x10

^{-6}
LIAL = 50m

**AREA EXPANSIVITY**

Area expansivity or superficial expansivity is defined as
the increase in area, per unit rise in temperature.

B = A

_{2}-A_{1}=Î”A
A

_{1}(Î¸_{2}Î¸_{1}) A1Î”Î¸
Where B = co-efficient of area expansivity

NOK: B = 2Î± i.e. Area expansivity is twice linear
expansivity.

B = Beta

Simplified 33 the length of a solid metallic cube at 200k is
5.0m. Give that the co-efficient of linear expansivity of the metal is 4.0x10

^{-5 }find the area of the cube at 120^{0}k.
Solution

A

_{1}= 5.0m, 01 = 20^{0}k
A

_{2}=? Î¸_{2}= 120^{0}k, Î± = 4.0x10^{-5}
But B = 2 Î± B = 2x4

^{0}0x10^{-5}=8x10^{-9}
B =
A

_{2}– A_{1}make A_{2}subject
A

_{1}(Î¸_{2}–Î¸_{1}) formula
A

_{2}= BA_{1}(Î¸_{2}–Î¸_{1})xA_{1}
A

_{2}= 4x10^{-4}x 100 +5 = 5.05m**VOLUME OR CUBIC EXPANSIVITY**

This is defined as the increase in volume of a substance per
unit volume per degree rise in temperature

ï»»
= Î”V = V

_{2}-V_{1}
V

_{1}Î”Î¸ V_{1}(Î¸_{2}–Î¸_{1})
But ï»¹=3
Î±

ï»¹
= co-efficient of volume expansivity

Simplified 34 calculate the change in volume when 1500cm3 of
steel is healed from 0

^{0}k to 40^{0}k (co-efficient of linear expansivity = 1.2x10+k^{-1})
Solution

ï»¹
= Î”V but ï»¹
= 3Î±

V1Î”Î¸ ï»¹=3x1.2x10

^{-5}k^{-1}
= 3.6x10

^{-5}
= 3.6x10

^{-5}= Î”V
1500 x 40

Î”V = 36x10

^{-5}x 60,000 = 2.16cm^{3}
Simplified 35 the volume of a metallic cube of 2

^{0}0k is 5.0cm given that the co-efficient of linear expansivity of the metal is 4.0x10^{-5}k^{-1}
Find the volume of the cube at 12

^{0}0k
Solution

V

_{1}= 5.0x5.0x5.0 = 125cm^{3}
ï»¹
= 3Î± = 3x4.0x10

^{-5}= 12x10^{-5}
Î¸

_{2}–Î¸_{1 }= 120-20 =100^{0}k
=ï»¹ = V

_{2}– V_{1}
V

_{1}Î¸_{2}–Î¸_{1}
12 x10

^{-5 }= V2 -125 = V_{2}-125
125x100 12500

V

_{2}– 125 = 125 = V_{2}– 125
V

_{2}= 1.5 +125 = 126.5cm3**EXPANSION IN LIQUID**

All liquids expand when heated and contracts on cooling,
different liquid leave different expansion rate. Since liquids have no length
or surface area and only increase in volume, only their volume or cubic
expansively can be discussed. But because the expansion of liquids is
complicated by the expansion of the containers. It is necessary to distinguish
between the real and apparent cubic expansivity of a liquid REAL (absolute)
cubic expansivity (Y1) of a liquid is the increase in volume per unit volume
per degree rise in temperature. Apparent (cubic expansivity (Ya) of a liquid is
the increase in volume [er unit volume per degree rise in temperature when the
liquid is heated in an expansible vessel.

Hence Y

_{1}= Y_{a}x Y_{c}
where Y1 = Real expansion

Y

_{a}= apparent expansion
Y

_{e}= cubic expansion
Y

_{c}= cubic expansion
Ya = loss in volume of liquid

Original
volume of liquid x temperature change

OR

Ya = loss in mass of liquid

Original mass of liquid x temperature change

Ya =
V

_{2}– V_{1}= M_{2}–M_{1}
V

_{1}x(Î¸_{2}–Î¸_{1}) M_{1}x (Î¸_{2}–Î¸_{1})
Simplified 36 a density glass bottle contains 44.25g of a
liquid at 0

^{0}c and 42.42g at 50^{0}c calculate the real cubic expansivity of the liquid (linear expansivity of glass = 1.0x10^{-5}k^{-1}
Solution

M

_{1}= 44.25g
Î¸

_{1}= 0^{0}c
M

_{2}= 42-02g
Î¸

_{2}= 50^{0}C
Î±glass = 1.0x10

^{-5}
Y

_{1}= ?
Y

_{1}= Y_{a}– Y_{c}
Y

_{c}= 3Î±
Y

_{c}= 3x1.0x10-5
Y

_{c}= 3x10-5
Y

_{2}= M_{1}– M_{2}
M

_{1}x (Î¸_{2}–Î¸_{1})
Y

_{2}= 44.25 – 42.02
44.25 x (50-0)

= 1.0077 x 10

^{-5}
Y

_{1}= 1.037x10^{-5}k^{-1}**ANOMALOUS EXPANSION OF WATER**

Mostly all liquids expands or increase in their volume
content where their temperature changes as a result of application of heat
water behaves irregularly or abnormal between 0

^{0}c & 4^{0}C this abnormal behave is called the Anomalous expansion of water it is a case were water contacts in volume between 0^{0}c & 40c and afterwards expands in volume like any other liquid between 0^{0}c and 4^{0}c, the volume of water reduces thereby increasing its density. Water has the minimum volume at 4^{0}c and maximum density at 4^{0}c this behavior of water when heat is applied is shown in the graph below.
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