Animated Solution for Physics - Thermodynamics: A metal of mass $1\text{ kg}$ at constant atmospheric pressure and at initial temperature $20^{\circ }\text{C}$ is given a heat of $20000\text{ J}$. Find the following : (a) change in temperature, (b) work done and (c) change in internal energy. (Given, Specific heat $=400{\text{ J/kg/}}^{\circ }\text{C}$, coefficient of cubical expansion, $\gamma =9\times 10^{-5}{/}^{\circ }\text{C}$, density $\rho =9000{\text{ kg/m}}^3$, atmospheric pressure $=10^5{\text{ N/m}}^2$)

The problem of heating a solid block of metal might seem trivial at first glance, but it beautifully illustrates the stark contrast between how solids and gases respond to heat. When you pump energy into a gas, it expands dramatically, doing significant work against its surroundings. But what happens when you do the same to a rigid piece of metal? Let's dive into the thermodynamics of solids and uncover the story hidden within the numbers.

Heating a Solid

Imagine a solid block of metal, with a mass of exactly $1\text{ kg}$, resting on a table. It is initially at a comfortable room temperature of $20^{\circ }\text{C}$. The atmosphere presses down on it with a constant pressure of $10^5{\text{ N/m}}^2$.

Now, we introduce a massive amount of energy: $20000\text{ J}$ of heat ($Q$) is pumped directly into the metal. Our goal is to track where every single Joule of this energy goes. Does it make the metal hotter? Does it make it expand? Let's find out.

The Temperature Rise

The most obvious effect of adding heat is a rise in temperature. The relationship between heat added and temperature change is governed by the specific heat capacity ($s$) of the material. The formula is beautifully simple:

We know the heat supplied $Q=20000\text{ J}$, the mass $m=1\text{ kg}$, and the specific heat $s=400{\text{ J/kg/}}^{\circ }\text{C}$. Rearranging the formula to solve for the change in temperature ($\Delta T$), we get:

Substituting our known values:

The metal's temperature increases by exactly $50^{\circ }\text{C}$. If it started at $20^{\circ }\text{C}$, it is now at a scorching $70^{\circ }\text{C}$.

The Work Done Against the Atmosphere

As the metal heats up, its atoms vibrate more vigorously, pushing each other slightly apart. This causes the entire block to expand. Because it is expanding against the constant atmospheric pressure, it must do work. The work done ($W$) at constant pressure is given by:

To find the work, we first need the change in volume ($\Delta V$). We are given the coefficient of cubical expansion ($\gamma =9\times 10^{-5}{/}^{\circ }\text{C}$). The formula for volume expansion is:

But wait, we don't have the initial volume ($V$). However, we do have the mass ($m=1\text{ kg}$) and the density ($\rho =9000{\text{ kg/m}}^3$). Since density is mass per unit volume, we can write $V=\frac{m}{\rho }$. Substituting this into our expansion formula:

Now, let's plug in the numbers:

This is an incredibly tiny expansion! Now, we can calculate the work done:

The metal does a mere $0.05\text{ J}$ of work pushing the atmosphere away.

The First Law and Internal Energy

Finally, we turn to the grand accountant of the universe: The First Law of Thermodynamics. It states that the heat supplied to a system must equal the change in its internal energy ($\Delta U$) plus the work done by the system:

We want to find the change in internal energy, so we rearrange the equation:

We supplied a massive $20000\text{ J}$ of heat, and the metal only spent $0.05\text{ J}$ doing work.

The Grand Conclusion: This result is profound. When you heat a solid, almost $100\%$ of the energy goes directly into increasing its internal energy (making the atoms vibrate faster). Because solids are so rigid and expand so little, the energy wasted on doing work against the atmosphere is practically zero. This is why, for solids and liquids, the specific heat at constant pressure ($C_p$) and the specific heat at constant volume ($C_v$) are nearly identical!