Question

A copper cylinder is initially at 20.0ºC. At what temperature will its volume be 0.150% larger than it is at 20.0ºC?

Short answer

The Temperature at which the volume increases by 0.150% is$49.4^0\mathrm{C}$ .

Step-by-step solution

Identification of given data

Initial temperature of copper cylinder is ${\mathrm{T}}_{\mathrm{i}}=20.0{}^0\mathrm{C}$

For copper, the coefficient of volume expansion$\beta =5.1\times {10}^{-5}{\left({}^{0}C\right)}^{-1}$

Percentage change in volume of the cylinder is given as$\frac{∆\mathrm{V}}{{\mathrm{V}}_0}=0.150\%$

Concept of Volume Expansion in solids.

Volume expansion is the expansion of the dimensions of an object with respect to temperature changes and can be expressed as,

$\mathbf{∆}\mathbf{V}\mathbf{=}{\mathbf{V}}_{\mathbf{0}}\mathbf{\beta }\mathbf{∆}\mathbf{T}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, V is the expansion in volume, V₀ is the unexpanded volume, β is the coefficient of volume expansion and T is the absolute temperature.

 Determination of the diameter value of the rivet following the conditions mentioned.

For copper, the coefficient of volume expansion,

$\mathrm{\beta }=5.1\times {10}^{-5}{\left({\mathrm{C}}^0\right)}^{-1}$

The percentage change in volume of the cylinder is given to be 0.150%.

So, $\frac{\mathrm{\Delta V}}{{\mathrm{V}}_0}=0.150\times {10}^{-2}$

Rearrange equation (i) to get the temperature change,

$$\begin{gathered} ∆\mathrm{T}=\frac{∆\mathrm{V}}{{\mathrm{V}}_0\mathrm{\beta }} \\ =\frac{0.150\times {10}^{-2}}{5.1\times {10}^{-5}{{(}^0\mathrm{C})}^{-1}} \\ =29.4{}^0\mathrm{C} \end{gathered}$$

Thus, the temperature at which the volume expanded 0.150% is,

$$\begin{gathered} {\mathrm{T}}_{\mathrm{f}}={\mathrm{T}}_{\mathrm{i}}+∆\mathrm{T} \\ =20.0^0\mathrm{C}+29.4^0\mathrm{C} \\ =49.4^0\mathrm{C} \end{gathered}$$