# Thermal expansion question and answer

Thermal expansion: solid or liquid expand when heat is applied to them. The expansion of any material both liquid and solid depend on the nature of the substance.

# Thermal expansion question and answer 1

Define the followings

(a),, (b). Cubic or expansivity a nd  (c). Superficial expansivity

(a) Linear expansity os solid can be defined as  increase in length per unit length, per unity rise  in temperature. It is given as

$\alpha \quad =\frac { { L }_{ 2 }{ -L }_{ 1 } }{ { L}_{ 1 }{ ({ \theta }_{ 2 }{ -\theta }_{ 1 }) } } { k }^{ -1 }$

(b). Cubic or volume expansivity, is defined as the increase in volume of a liquid per a unit volume per unit rise in temperature. It is given as

$\gamma \quad =\frac { { V }_{ 2 }{ -V }_{ 1 } }{ { V }_{ 1 }{ ({ \theta }_{ 2 }{ -\theta }_{ 1 }) } }$

(c). Superficial of area expansivity is defined as increase in area of a substance per unit area per unit rise in temperature. The formula is also given as

$\beta \quad =\frac { { A }_{ 2 }{ -A }_{ 1 } }{ { A }_{ 1 }{ ({ \theta }_{ 2 }{ -\theta }_{ 1 }) } }$

# thermal expansion question and answer  2

A brass of length 100m increases to 100.5m, when heated from 50oc to 100oc. Calculate it linear expansivity

$\alpha =\frac { { L }_{ 2 }{ -L }_{ 1 } }{ { L }_{ 1 }({ \theta }_{ 2 }-{ \theta }_{ 1 }) }$

But  (Ѳ21) =  100oc – 50oc = 50oc,  L1-L2 = 100.5 = 100 = 0.5m, L1 = 100m, and α= ?

$\alpha =\frac { 100.5-100 }{ 100(100-50) } =\frac { 0.5 }{ 100\times 50 } \quad =\frac { 0.5 }{ 5000 } =1\times { 10 }^{ -4 }$

Linear expansivity α = 10-4K-1

# thermal expansion question and answer 3

An iron rod of linear expansivity 24 x 10-6K-1 and L1 = 60m, expands when heated through 100oc, calculate (a) increase in length and (b) final length.

$\alpha =\frac { { L }_{ 2 }{ -L }_{ 1 } }{ { L }_{ 1 }({ \theta }_{ 2 }-{ \theta }_{ 1 }) }$

${ L }_{ 2 }{ -L }_{ 1 }=\alpha { L }_{ 1 }\theta$

L1 =60m, Ѳ=100oc, α = 12 x 10-6K-1 and ( L2-L1 )=?

L2-L1 = 12 x10-6 x60 x 100 = 0.072

The increase in length = 0.072m

(b). Final length = increase in length  + original length , that is

60 + 0.072 = 60.072m

# Thermal expansion question and answer 4

Mercury in glass thermometer has a bulb of volume 0.4cm3 and a tube of 20 x 10-5cm2. Calculate

(a). the apparent increase in volume of the mercury when the temperature rises from 0oc to 100oc

(b). the distance between the fixed points

${ (a)\quad \quad \gamma }_{ a }=\frac { { V }_{ 2 }{ -V }_{ 1 } }{ { V }_{ 1 }({ \theta }_{ 2 }{ \theta }_{ 1 }) }$

V2-V1 = γaV121) where V1= 0.4cm, (Ѳ2-Ѳ1) = 100oc -0oc = 100oc, γa = 12 x 10-6c-1

V2-V1 = 12 x 10-6 x 0.4 x 100 = 48 x 10-5

Increase in volume = 48 x 10-5cm3

(b). to calculate the distance between fixed point, consider calculating the volume of the tube that is

20 x 10-5cm2 x h = 48 x 10-5

0.00002h = 0.00048

$h=\frac { 0.00048 }{ 0.0002 } \quad =\quad 2.4cm$

The distance between the fixed point = 2.4cm

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