Question

A brass ring of diameter 10.00 cm at 19.0°C is heated and slipped over an aluminum rod with a diameter of 10.01 cm at 19.0°C. Assuming the average coefficients of linear expansion are constant. What if the aluminum rod were 10.02 cm in diameter?

Answer 1

Answer:

the final temperature is $\mathbf{T_f = -377.2^0 C}$

Explanation:

The change in length of a bar can be expressed with the relation;

$\Delta L = L_f - L_i$   ---- (1)

Also ; the relative or fractional increase in length  is proportional to the change in temperature.

Mathematically;

ΔL/L_i ∝ kΔT

where;

k is replaced with ∝ (the proportionality constant )

$\dfrac{ \Delta L}{L_i}=\alpha \Delta T$    ---- (2)

From (1) ;

$L_f = \Delta L + L_i$  ---  (3)

from (2)

${ \Delta L}=\alpha \Delta T*{L_i}$  ---- (4)

replacing (4) into (3);we have;

$L_f =(\alpha \Delta T*{L_i} ) + L_i$

On re-arrangement; we have

$L_f = L_i + \alpha L_i (\Delta T )$

from the given question; we can say that :

$(L_f)_{brass}}} = (L_f)_{Al}$

So;

$L_{brass} + \alpha _{brass} L_{brass}(\Delta T) = L_{Al} + \alpha _{Al} L_{Al}(\Delta T)$

Making the change in temperature the subject of the formula; we have:

$\Delta T = \dfrac{L_{Al}-L_{brass}}{\alpha _ {brass} L_{brass}-\alpha _{Al}L_{Al}}$

where;

$L_{Al}$ = 10.02 cm

$L_{brass}$ = 10.00 cm

$\alpha _{brass}$ = 19 × 10⁻⁶ °C ⁻¹

$\alpha_{Al}$ = 24  × 10⁻⁶ °C ⁻¹

$\Delta T = \dfrac{10.02-10.00}{19*10^{-6} \ \ {^0}C^{-1} *10.00 -24*10^{-6} \ \ {^0}C^{-1} *10.02}$

$\Delta T$ = −396.1965135 ° C

$\Delta T$ ≅ −396.20  °C

Given that the initial temperature $T_i = 19^0 C$

Then ;

$\Delta T = T_f - T_i$

$T_f = \Delta T + T_I$

Thus;

$T_f =(-396.20 + 19.0)^0 C$

$\mathbf{T_f = -377.2^0 C}$

Thus; the final temperature is $\mathbf{T_f = -377.2^0 C}$