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 (α) = L₂-L₁

                    L1 x (ө₂- ө₁)

Where

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?

Solution

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)

Solution

L₁ = 90m, or = 00k, L₂? Θ2 = 55⁰k

α = L₂-L₁      make L₂ subject

     L₁ θ₂-θ₁ formula

 L₂ = α L₁ (θ₂-θ₁) + 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

Solution

L₁ = 3m, safety gap = Δ2 = ? θ₁ = 28⁰k

θ₂ = 400k, α = 1.0x10^-3k^-1

α = ΔL

     L₁Δθ

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

Solution

L₁ =40m, L₂ = 40 05, θ₁ = 20⁰k,θ₂ =45⁰k

α =?

α = L2 – L1 = 40-05-40

       L₁θ₂-θ₁    40x45-20

= 0.005 = 5x10^-5k^-1

    1000

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)

Solution 

For iron

α fe = ΔLfe

       L₁ Δθfe

ΔLfe =αfeL₁Δθfe

For aluminum

αAL = ΔLAL

         L₁ ΔθAL

ΔLAL =α AL L₁ AθAL

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

24x10^-6

LIAL = 50m

AREA EXPANSIVITY

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

B = A₂-A₁            =ΔA

     A₁ (θ₂ 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.

Solution

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

       A₁ (θ₂ –θ₁) formula

A₂ = BA₁ (θ₂ –θ₁)xA₁

A₂ = 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

ﻻ = ΔV = V₂-V₁

V₁Δθ      V₁ (θ₂ –θ₁)

But ﻹ=3 α

ﻹ = 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)

Solution

ﻹ = ΔV but ﻹ = 3α

V1Δθ  ﻹ=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

Solution

V₁ = 5.0x5.0x5.0 = 125cm³

ﻹ = 3α = 3x4.0x10^-5 = 12x10^-5

θ₂ –θ₁ = 120-20 =100⁰k

=ﻹ = V₂ – V₁

        V₁θ₂ –θ₁

12 x10^-5 = V2 -125   =  V₂ -125

                125x100       12500

V₂ – 125 = 125 =  V₂ – 125

 V₂ = 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 Y₁ = 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 =    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

Solution

M₁ = 44.25g

θ₁ = 0⁰c

M₂ = 42-02g

θ₂ = 50⁰C

αglass = 1.0x10^-5

Y₁ = ?

Y₁ = Y_a – Y_c

Y_c = 3α

Y_c = 3x1.0x10-5

Y_c = 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

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 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.