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

Answer

(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 }) } }

Question an answer on momentum

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

Answer

\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

Question an answer on  elasticity 

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.

Answer

\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

Answer

{ (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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