Animated Solution for Physics - Properties of Solids and Liquids: Each side of a box made of metal sheet in cubic shape is $a$ at room temperature $T$, the coefficient of linear expansion of the metal sheet is $\alpha$. The metal sheet is heated uniformly, by a small temperature $\Delta T$, so that its new temperature is $T+\Delta T$. Calculate the increase in the volume of the metal box.

Have you ever wondered why bridges have expansion joints or why a tightly sealed metal lid opens easily after being run under hot water? The answer lies in the fascinating phenomenon of thermal expansion.

When a material is heated, its constituent atoms and molecules gain kinetic energy. They begin to vibrate more vigorously around their equilibrium positions. Because the interatomic forces are slightly asymmetric, these increased vibrations push the atoms further apart on average. Macroscopically, this results in the entire object expanding in all directions. In this problem, we are tasked with calculating the exact increase in the volume of a hollow metal cube when it is subjected to a temperature rise.

We are given a box made of metal sheets, forming a perfect cube. At room temperature $T$, each side of this cube has a length of $a$. The material has a coefficient of linear expansion denoted by $\alpha$.

The box is then heated uniformly by a small temperature difference, $\Delta T$. Our goal is to find the increase in the volume of the box, which we denote as $\Delta V$.

To find the change in volume, we rely on the fundamental formula for volumetric thermal expansion:

Here, $V$ is the original volume of the object, $\gamma$ is the coefficient of volume expansion, and $\Delta T$ is the change in temperature.

However, there is a slight catch. The problem provides us with $\alpha$ (the linear expansion coefficient), not $\gamma$. How do we bridge this gap?

For an isotropic material—a material that expands equally in all directions—the volume expansion coefficient is simply three times the linear expansion coefficient.

Why is this the case? Imagine a cube of side $L$. When heated, the new side length becomes $L^{\prime }=L(1+\alpha \Delta T)$. The new volume is $V^{\prime }=(L^{\prime }{)}^3=L^3(1+\alpha \Delta T{)}^3$. If we expand this using the binomial theorem, we get $V^{\prime }=V(1+3\alpha \Delta T+3{\alpha }^2\Delta T^2+{\alpha }^3\Delta T^3)$. Because $\alpha$ is an incredibly small number (typically on the order of $10^{-5}$), the terms containing ${\alpha }^2$ and ${\alpha }^3$ are vanishingly small and can be safely ignored. This leaves us with $V^{\prime }\approx V(1+3\alpha \Delta T)$, proving that the effective volume expansion coefficient is indeed $3\alpha$.

You might be wondering: "The question explicitly states that the box is made of metal sheets. It's hollow! Does a hollow box expand differently than a solid block?"

This is a classic conceptual trap in physics. The beautiful truth is that a hollow object expands exactly as if it were a solid block of the same material. Think of thermal expansion as a photographic enlargement. Every single dimension, including the empty space inside the cavity, scales up by the exact same factor. Therefore, the volume enclosed by the metal sheets expands volumetrically just like a solid cube would.

Now that we have all our conceptual tools, the execution is straightforward.

First, we determine the initial volume of the cube. Since the side length is $a$, the volume is:

Next, we substitute $V=a^3$ and $\gamma =3\alpha$ into our master equation for volumetric expansion:

Rearranging the terms for standard mathematical elegance, we arrive at our final answer:

And there we have it! By understanding the relationship between linear and volumetric expansion, and recognizing that cavities expand just like solid material, we've effortlessly solved the problem.