Question

Solve the following problem.

An aluminium rod and iron rod show 1.5 m difference in their lengths when heated at all temperature. What are their lengths at 0 °C if coefficient of linear expansion for aluminium is 24.5 × 10^–6/°C and for iron is 11.9 × 10^–6/°C?

Answer 1

Given: (L_T)_i - (L_T)_al = 1.5 m, T₀ = 0 °C,
α_al = 24.5 × 10^–6/°C
α_i = 11.9 × 10^–6/°C

To find: Lengths of aluminium and iron rod (L₀)_al and (L₀)_i

Formula: L_T = L₀ [(1 + α (T - T₀)]

Calculation: For T₀ = 0 °C
From formula, L_T = L₀ (1+ αT)
For aluminium,
(L_T)_al = (L₀)_al (1+ α_al T)       ….(1)
For iron,
(L_T)_i = (L₀)_i (1 + α_iT)     ….(2)

Subtracting equation (2) by (1),

(L_T)_i = (L_T)_al = [(L₀)_i + (L₀)_i α_iT] - [(L₀)_al + (L₀)_al  α_al T]

= (L₀)_i - (L₀)_al + [(L₀)_i α_i - (L₀)_al α_al ]T

∴ 1.5 = 1.5 + [(L₀)_i α_i - (L₀)_al α_al] T

⇒ [(L₀)_i α_i - (L₀)_al α_al] T = 0

∴ (L₀)_al α_al = (L₀)_i α_i 

∴ (L₀)_al = (L₀)_i α_i 

∴ (L₀)_al = (L₀)_i `alpha_"i"/alpha_"al"`

`= ("L"_0)_"i" xx (11.9 xx 10^-6)/(24.5 xx 10^-6)`

`= ("L"_0)_"i" xx 17/35`

`("L"_0)_"al" = [("L"_0)_"al" + 1.5] xx 17/35  .....["Given:" ("L"_"T")_"i" - ("L"_"T")_"sl" = 1.5 "m"]` 

∴ `35 ("L"_0)_"al" = 17 ("L"_0)_"al" + 1.5 xx 17`

∴ `35 ("L"_0)_"al" - 17 ("L"_0)_"al" = 1.5 xx 17`

∴ `18 ("L"_0)_"al" = 1.5 xx 17`

∴ `("L"_0)_"al" = (1.5 xx 17)/18` = 1.417 m

∴ `("L"_0)_"i" = 1.5 + ("L"_0)_"al"`

= 1.5 + 1.417

= 2.917 m

Length of aluminium rod at 0 ºC is 1.417 m and that of iron rod is 2.917 m.