Question

An aluminum sphere is 9.00 cm in diameter

What will be its % change in volume if it is heated from 30∘C to 180 ∘C?

Answer 1

Given,

d = 9 cm ; T1 = 30 deg C = 303 k ; T2 = 180 deg C = 453 K

D = Do ( 1 + $\alpha$$\Delta$T)

D = 9 [ 1 + x 22.2 x 10^-6 x (453 - 303) ] = 9.03 cm

R = 9.03 cm/2 = 4.52 cm

V = 4/3 pi r³ = 4/3 x 3.14 x 4.52³ = 386.82 cm³

Vo =  4/3 pi ro³ = 4/3 x 3.14 x 4.5³ = 381.7 cm³

% change in volume

%$\Delta$V = ( 386.82 - 381.7) / 381.7 = 1.34

Percentage change of 1.34 %.