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

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

Step-by-step 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 )

`( \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 = \frac{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 = \frac{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}`