Question

A cube of solid aluminium has a volume of $1m^3$at ${25}^{o}C$. What temperature change is required to produce a$100c{m}^3$ increase in the volume of the cube? [${\alpha }_{aluminium}=24\times {10}^{-6}{/}^{o}C$]

Answer 1

Step 1: Model the problem as Thermal volume expansion due to the rise in temperature-

1. We will consider expansion in volume as the rise in temperature.
2. We know that if $V$ is the final volume of the object after the temperature rise and $V_0$ is the initial volume of the object before temperature rise, then the relationship between the expansion coefficient, temperature rise and volumes can be written as $V=V_0(1+\gamma \Delta T)$

Step 2: Extract the given data from the problem.

1. The linear expansion coefficient of aluminium is given as ${\alpha }_{aluminium}=24\times {10}^{-6}/{}^{o}C$.
2. Temperature difference can be calculated as $\Delta T=T-25$, where $T$ is the final temperature and ${25}^{o}C$ is the initial temperature.
3. The initial volume of the cube $V_0=1m^3={10}^{6}c{m}^3$
4. we want to produce a $100c{m}^3$ of increase in the volume of the cube.

$\therefore V=\left(V_0+100\right)c{m}^3$

$\Rightarrow V-V_0=100c{m}^3$.

We also know that $\gamma =3\alpha$.

Step 3: Apply the linear thermal expansion formula using the above data-

$V=V_0(1+\gamma \Delta T)$

or, $V-V_0=3\alpha \Delta T{V}_0$

Putting values in the above equation,

$100=\left(3\times 24\times {10}^{-6}\right)\left(\Delta T\right)\left({10}^6\right)$

or, $\frac{100}{72}=\Delta T$

or, $\Delta T={1.39}^{o}C$

Therefore, we need to produce$1.{39}^{o}C$temperature change to produce a $100c{m}^3$of increase in the volume of the cube.