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

(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

(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

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# thermal expansion question and answer 2

**A brass of length 100m increases to 100.5m, when heated from 50 ^{o}c to 100^{o}c. Calculate it linear expansivity**

# Answer

But (Ѳ_{2}-Ѳ_{1}) = 100^{o}c – 50^{o}c = 50^{o}c, L_{1}-L_{2} = 100.5 = 100 = 0.5m, L_{1} = 100m, and α= ?

Linear expansivity α = 10^{-4}K^{-1}

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**thermal expansion question and answer 3**

**An iron rod of linear expansivity 24 x 10 ^{-6}K^{-1} and L_{1} = 60m, expands when heated through 100^{o}c, calculate (a) increase in length and (b) final length.**

# Answer

L1 =60m, Ѳ=100^{o}c, α = 12 x 10^{-6}K^{-1} and ( L_{2}-L_{1} )=?

L_{2}-L_{1} = 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 0.4cm ^{3} and a tube of 20 x 10^{-5}cm^{2}. Calculate**

**(a). the apparent increase in volume of the mercury when the temperature rises from 0 ^{o}c to 100^{o}c**

**(b). the distance between the fixed points**

# Answer

**V _{2}-V_{1} = γ_{a}V_{1}(Ѳ_{2}-Ѳ_{1}) **where V1= 0.4cm, (Ѳ2-Ѳ1) = 100

^{o}c -0

^{o}c = 100

^{o}c, γ

_{a}= 12 x 10

^{-6}c

^{-1}

**V ^{2}-V^{1} = 12 x 10^{-6} x 0.4 x 100 = 48 x 10^{-5}**

Increase in volume = 48 x 10^{-5}cm^{3}

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

20 x 10^{-5}cm^{2} x h = 48 x 10^{-5}

0.00002h = 0.00048

The distance between the fixed point = 2.4cm

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