# The temperature of an aluminum disk is increased by 250^{\circ}C . By what percentage does its...

## Question:

The temperature of an aluminum disk is increased by {eq}250^{\circ}C {/eq}. By what percentage does its volume increase?

## Volume Expansion:

Every metal has the property of volume expansion, and the coefficient of volume expansion for different meta is different. On the rise in the temperature, metal will expand volumetrically; this phenomenon is known as the volume expansion.

Given data

• The increase in temperature is: {eq}\Delta T = 250^\circ \;{\rm{C}} {/eq}

Expression for the volume expansion or change in volume is,

{eq}\Delta V = \beta {V_0}\Delta T {/eq}

The fractional increase in the volume is equal to the increase in the volume of disk divided by the initial volume disk, so rewrite the above equation,

{eq}\dfrac{{\Delta V}}{{{V_0}}} = \beta \Delta T {/eq}

To calculate percentage increase in the volume of the disk, multiply both sides with 100,

{eq}\left( {\dfrac{{\Delta V}}{{{V_0}}}} \right)100 = \left( {\beta \Delta T} \right)100\;\% {/eq}

Here, {eq}\beta = 69 \times {10^{ - 6}}\;{\rm{^\circ }}{{\rm{C}}^{{\rm{ - 1}}}} {/eq} is the coefficient volume expansion, {eq}{V_0} {/eq} is initial volume and {eq}\Delta V {/eq} change in the volume.

Substituting all the value in the above equation,

{eq}\begin{align*} \left( {\dfrac{{\Delta V}}{{{V_0}}}} \right)100 &= \left( {69 \times {{10}^{ - 6}} \times 250} \right)100\;\% \\ &= 1.725\;\% \end{align*} {/eq}

Thus, the increase in the volume is {eq}1.725\;\% {/eq}