The coefficient of linear expansion,
α_L =24 K^-1
The coefficient of cubical expansion,
α_v =3α_L= 3 x 24
= 72 K^-1
The initial temperature, T₁ = 303 K
The initial volume, V₀ = 0.3 m³
The final volume, V₁ = 0.35 m³
Change in volume ΔV = 0.35 - 0.30
= 0.05 m³
To find: The final temperature T₂= ?
ΔV = α_v. V₀ (ΔT)
0.05 = 72 x 0.3 x (T₂ - 303)
0.05 = 21.6 (T₂ - 303)
= 21.6 T₂ - 6544.8
21.6 T₂ = 6544.8 - 0.05
T₂ =\(\frac { 6544.8 }{ 21.6 } \)
Final temperature, T₂ = 303K

Initial temperature, t₁ = 285 k
Final temperature, t₂ = 363 k
Initial length, I₁ = 6.522 m
Change in length, Δl = 0.576 m
To find: Coefficient of linear expansion α_L=?
α_L =\(\frac { \triangle l }{ l\triangle T } =\frac { ({ l }_{ 2 }-{ l }_{ 1 }) }{ { l }_{ 1 }(t_{ 2 }-{ t }_{ 1 }) } \)
=\(\frac { 6.576-6.522 }{ 6.5222(63-285) } \)
=\(\frac { 0.054 }{ 6.522(78) } =\frac { 0.054 }{ 508.7 } \)
α_L =1.06 x 10^-4 K^-1.

Length of AI sheet, L = 30 m
Coefficient of linear expansion, Δl = 23 x 10^-6 K^-1
Change in temperature ΔT = 300 k
To find: Change in length ΔL =?
ΔL = L α_L ΔL
= 30 x 23 x 10^-6 x 300
= 2.07 x 10^-1
ΔL = 0.207 m

F to (kelvin) K =(F--32) x \(\frac{5}{9}\) + 273
=(95 - 32) x \(\frac{5}{9}\) + 273
=\(\frac { 315 }{ 9 } \) + 273
=35+273 =308 K

Coefficient of linear expansion α_L=\(\frac { \triangle L }{ { L } } \times \frac { 1 }{ \triangle T } \)
Length of a steel piece L =2m
Initial temperature (T_i) = 200 k
Final temperature (T_f) = 250 k
ΔT=T_f - T_i
Increase in temperature, ΔT = 250 -200
=50 k
Increase in length, ΔL = 0.1 m
To find: Coefficient of increase expansion α_L?
α_L =\(\frac { 0.1 }{ 2 } \times \frac { 1 }{ 50 }\)
α_L =0.001 k-1
Coefficient cubical expansion
= 3 x coefficient linear expansion