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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.

Change in temperature
Change in state
Emission of electrons
Resistance change

Solids expands when heated and contracts when cooled
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.

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 (α) = L2-L1
                    L1 x (ө2- ө1)
α – co-efficient of linear expansively the 5.1 unit is (k-1) per Kelvin Initre
L1 = length of the metal
L2 = final length of the metal

Q1 = initial temperature
Q2 = fuel temperature
NOTE: C2 – L1 = ΔL
Q2 – Q1 = Δθ
Simplified 30 the linear expansively of a metal is o.000011 park what does this statement mean?
It means that a unit substance of the metal will expand by 0.000011 when it’s temperature changes by 10K
Simplified 30 a telephone steel wire is 90m long at 00K. how much longer will it be at 550k (linear expansively of steel is 0.000011k-1)

L1 = 90m, or = 00k, L2? Θ2 = 550k
α = L2-L1      make L2 subject
     L1 θ21 formula
 L2 = α L121) + L1
L2 = 0.000011 x 90 (55-0) x 90
L2 = 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 1-0x10-5k-1. What is the safety gap that must be left between successive bars if the highest temperature expected is 400k

L1 = 3m, safety gap = Δ2 = ? θ1 = 280k
θ2 = 400k, α = 1.0x10-3k-1
α = ΔL
ΔL =α L1Δθ
= > 1.0x10-5 x 3x(40-28)
= 3.6x10-4m

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

L1 =40m, L2 = 40 05, θ1 = 200k,θ2 =450k
α =?
α = L2 – L1 = 40-05-40
       L1θ21    40x45-20
= 0.005 = 5x10-5k-1
Simplified 32 an iron rod is 100m at 100C what must be the length of an aluminum rod at 00c, if  the difference between length of the two rods is to remain the same at the same temperature (α fe = 12x10-6, αAl = 24x10-6)

For iron
α fe = ΔLfe
       L1 Δθfe
ΔLfe =αfeL1Δθfe
For aluminum
         L1 ΔθAL

But ΔLfe = ΔLAL
α fe L1 Δθ = αALL1Δθ
αfeL1 = α ALL1AL
life = 100m
αfe = 12x10-6
L1AL = ?
αAl = 24x10-6
 = 12 x10-6 x 100 = 24x10-6 x LINL
12x10-6 x 10 = LIAL
LIAL = 50m

Area expansivity or superficial expansivity is defined as the increase in area, per unit rise in temperature.
B = A2-A1            =ΔA
     A12 θ1)      A1Δθ
Where B = co-efficient of area expansivity
NOK: B = 2α 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 1200k.

A1 = 5.0m, 01 = 200k
A2 =? θ2 = 1200k, α = 4.0x10-5
But B = 2 α  B = 2x400x10-5 =8x10-9
B =   A2 – A1        make A2 subject
       A12 –θ1) formula
A2 = BA12 –θ1)xA1
A2 = 4x10-4 x 100 +5 = 5.05m

This is defined as the increase in volume of a substance per unit volume per degree rise in temperature
= ΔV = V2-V1
V1Δθ      V12 –θ1)
But =3 α
= co-efficient of volume expansivity
Simplified 34 calculate the change in volume when 1500cm3 of steel is healed from 00k to 400k (co-efficient of linear expansivity = 1.2x10+k-1)

= ΔV but = 3α
V1Δθ  =3x1.2x10-5k-1
= 3.6x10-5
= 3.6x10-5 =    ΔV
                   1500 x 40
ΔV = 36x10-5 x 60,000 = 2.16cm3
Simplified 35 the volume of a metallic cube of 200k is 5.0cm given that the co-efficient of linear expansivity of the metal is 4.0x10-5k-1
Find the volume of the cube at 1200k

V1 = 5.0x5.0x5.0 = 125cm3
= 3α = 3x4.0x10-5 = 12x10-5
θ2 –θ1 = 120-20 =1000k

= = V2 – V1
        V1θ2 –θ1
12 x10-5 = V2 -125   =  V2 -125
                125x100       12500

V2 – 125 = 125 =  V2 – 125
 V2 = 1.5 +125 = 126.5cm3
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 Y1 = Ya x Yc
where Y1 = Real expansion
Ya = apparent expansion
Ye = cubic expansion
Yc = cubic expansion
Ya = loss in volume of liquid
        Original volume of liquid x temperature change
Ya = loss in mass of liquid
Original mass of liquid x temperature change
Ya =    V2 – V1 =           M2 –M1
            V1x(θ2 –θ1)       M1 x (θ2 –θ1)
Simplified 36 a density glass bottle contains 44.25g of a liquid at 00c and 42.42g at 500c calculate the real cubic expansivity of the liquid (linear expansivity of glass = 1.0x10-5k-1
M1 = 44.25g
θ1 = 00c
M2 = 42-02g
θ2 = 500C
αglass = 1.0x10-5
Y1 = ?
Y1 = Ya – Yc
Yc = 3α
Yc = 3x1.0x10-5
Yc = 3x10-5
Y2 = M1 – M2
        M1 x (θ2 –θ1)
Y2 = 44.25 – 42.02
44.25 x (50-0)
= 1.0077 x 10-5
Y1 = 1.037x10-5k-1

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 00c & 40C this abnormal behave is called the Anomalous expansion of water it is a case were water contacts in volume between 00c & 40c and afterwards expands in volume like any other liquid between 00c and 40c, the volume of water reduces thereby increasing its density. Water has the minimum volume at 40c and maximum density at 40c this behavior of water when heat is applied is shown in the graph below.

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