Question

A metal plate of dimensions $20\;cm \times 10\;cm \times 0.1\;cm$ having $\alpha = 5\times 10^{-4}\; ^{\circ}C^{-1}$. Find the change in the area of the plate on changing the temperature by $50^{\circ} C$.

Answer 1

Recall the expression for the coefficient of thermal expansion with respect to the change in the dimensions of the metal plate. It relates the change in the metal plate’s area dimensions to a change in temperature. To this end, plug in the given values to arrive at the change in the area of the metal plate. Note that the coefficient of expansion given to us is for one dimensional evaluation. Convert it to dimensions of area before plugging it into your equation.

Formula Used:
Coefficient of (thermal) area expansion $\beta = \dfrac{1}{A}.\dfrac{\Delta A}{\Delta T}$

Complete answer:
Let us begin by listing out the parameters given to us.
We have a metal plate of length $l = 20\;cm = 0.2\;m$, breadth $b = 10\;cm = 0.1\;m$ and height $h = 0.1\;cm = 0.001\;m$.
The area of the metal plate will be $A = l \times b = 0.2 \times 0.1 = 0.02\;m = 2 \times 10^{-2}\;m$
We are given that the coefficient of linear expansion of the metal plate is $\alpha = 5\times 10^{-4}\; ^{\circ}C^{-1}$. Since we are concerned with the area of the metal plate which essentially entails two dimensions, we can convert the coefficient of linear expansion into that of area expansion $\beta$ by multiplying it by a factor of 2 to account for two dimensions of the metal plate, i.e.,
$\beta = 2\alpha = 2 \times 5 \times 10^{-4} = 10 \times 10^{-4} = 10^{-3}\; ^{\circ}C^{-1}$
We are also given that the temperature is changed by $\Delta T = 50^{\circ} C$.
Now, we know that the coefficient of area expansion is defined as the change in the area of an object for per degree change in its temperature., i.e.,
$\beta = \dfrac{1}{A}.\dfrac{\Delta A}{\Delta T}$
The change in area can be calculated as:
$\Delta A = \beta \times A \times \Delta T$
Plugging in our values we get:
$\Delta A = 10^{-3} \times 2 \times 10^{-2} \times 50 = 10^{-3-2+2}= 10^{-3}\;m^2 = 0.001\;m^2$
Thus, the area of the metal plate changes by $0.001\;m^2$ when the temperature is changed by $50^{\circ}C$.

Note:
We can mathematically derive the relation between the linear, area and volume coefficients of expansion as follows:
Let the linear coefficient of expansion be given as: $\alpha = \dfrac{1}{L}.\dfrac{d L}{d T}$, and let the area coefficient of expansion be: $\beta = \dfrac{1}{A}.\dfrac{d A}{d T}$
We know that area is more or less the linear dimensions squared, i.e., $A = L^2$
Taking logarithm on both sides:
$ln\;A = ln\;L^2 \Rightarrow ln\;A = 2\;ln\;L$
Differentiating the above equation on both sides with respect to T:
$\dfrac{d}{dT} ln\;A = 2\;\dfrac{d}{dT}ln\;L$
$\Rightarrow \dfrac{1}{A}.\dfrac{dA}{dT} = 2.\dfrac{1}{L}.\dfrac{dL}{dT} = 2.\alpha$
$\Rightarrow \beta = 2\alpha$
Similarly, solving for volume by taking $V = L^3$ also results in the volume coefficient of expansion to be $\gamma=3\alpha$.