Energy Conservation with Internal Energy

Hello! In our last lesson, we tackled complex problems involving pistons, springs, and friction, focusing on how a gas does work on its mechanical surroundings. We carefully calculated the work done by the gas, $W_{g a s}$ , and used it in the First Law, $Δ U = Q - W$ .

Today, we'll investigate the reverse energy flow, which is the very foundation of the First Law of Thermodynamics. Our goal is to apply conservation of energy to systems where macroscopic mechanical work is converted into internal energy. This occurs in many physical situations, from friction and inelastic collisions to the simple act of compressing a gas. Mastering this concept is key to solving integrated problems where mechanics and thermodynamics intersect.

1. The Mechanical Equivalent of Heat: Joule's Foundational Insight

Before the 1840s, heat and work were considered distinct concepts. James Prescott Joule's brilliant experiments demonstrated that they are simply different ways of transferring energy. He showed that a specific amount of mechanical work done on a system, such as stirring water with a paddle wheel, always produced a predictable temperature increase—the same increase that could be achieved by adding a specific amount of heat.

This established the mechanical equivalent of heat: work can be converted directly into internal energy.

To understand Joule's reasoning and how it defines internal energy as a state function, please read the following excerpt from the NCERT textbook.

Thermodynamics - NCERT

The NCERT textbook provides a clear historical and conceptual background for the First Law, starting with Joule's experiments. This section explains how performing work on an insulated (adiabatic) system leads to a change in its internal energy.

Please read the section 'The Internal Energy as a State Function', starting from subsection (a) 'Work'. Focus on how Joule's experiments showed that the change in the system's state (measured by temperature) depended only on the amount of work done, not on how the work was done. This established internal energy as a state function.

A crucial point to note from the text is the sign convention. The NCERT textbook, like many chemistry resources, uses the convention $Δ U = q + w$ , where $w$ is the work done on the system. As we've discussed, our course uses the physics/engineering convention $Δ U = Q - W$ , where $W$ is the work done by the system.

They are perfectly consistent because $W_{by system} = - W_{on system}$ . Joule's experiment, in our convention, is an adiabatic process ( $Q = 0$ ) where work is done on the system ( $W < 0$ ). The First Law gives:
$Δ U = Q - W = 0 - W = - W$
Since $W$ is negative, $Δ U$ is positive. The internal energy increases by an amount exactly equal to the work done on the system.

2. Case Studies: Converting Work to Internal Energy

Let's explore three common scenarios where this energy conversion is central.

Case 1: Adiabatic Compression

When you pump up a bicycle tire, the pump gets hot. This isn't just due to friction; you are doing work on the gas, and that work is being converted into internal energy.

System: The gas inside the pump.
Process: The piston is pushed, compressing the gas. If this happens quickly, there is little time for heat to escape, so we can approximate it as adiabatic ( $Q \approx 0$ ).
Analysis: Work is done on the gas, so $W < 0$ .
$Δ U = Q - W \approx 0 - W = - W$
Since $W$ is negative, $Δ U$ is positive. The mechanical work done to push the piston directly increases the internal energy of the gas, raising its temperature.

The following video calculates the work done during compression. While it uses the chemistry sign convention, it's an excellent exercise for you to translate the concepts and confirm the final energy transfer.

Internal Energy, Heat, and Work Thermodynamics, Pressure & Volume, Chemistry Problems

This video from The Organic Chemistry Tutor demonstrates how to calculate the work involved in compressing a gas. It's a great opportunity to practice your understanding of sign conventions.

Watch the segment from 18:27 to 19:58. The problem is to calculate the work required to compress a gas. Notice that the change in volume \Delta V is negative. Using the chemistry formula w = -P\Delta V, the work done on the system comes out positive, as expected. How would you calculate the work done by the system, W, in our convention?

In our convention, work done by the gas is $W = \int P d V$ . For constant pressure, $W = P Δ V$ . Since $Δ V$ is negative for compression, $W$ is negative, correctly indicating work was done on the system.

Case 2: Friction and Dissipation

Friction is the quintessential dissipative force. It converts ordered macroscopic kinetic energy into disordered microscopic thermal energy.

Consider a block of mass $m$ sliding with initial speed $v_{i}$ on a rough horizontal surface and coming to a stop.

System: Block + Surface.
Process: The block's kinetic energy is dissipated by the work done by friction.
Energy Conservation: The initial kinetic energy of the block is the only energy input. It's entirely converted into the internal energy of the block and the surface, causing their temperatures to rise.
$Δ K E + Δ U = 0$
$Δ U = - Δ K E = - (0 - \frac{1}{2} m v_{i}^{2}) = \frac{1}{2} m v_{i}^{2}$
The total increase in internal energy of the block-surface system equals the initial kinetic energy of the block.

Case 3: Inelastic Collisions

In a perfectly inelastic collision, two objects stick together, and macroscopic kinetic energy is not conserved. This "lost" kinetic energy is converted into internal energy, resulting in deformation and a temperature increase.

Imagine a bullet of mass $m$ and velocity $v$ striking and embedding itself in a stationary wooden block of mass $M$ .

Mechanics (Momentum Conservation): First, find the final velocity $v_{f}$ of the combined system.
$p_{i} = p_{f} ⟹ m v = (m + M) v_{f} ⟹ v_{f} = \frac{m}{m + M} v$
Energy (Calculate KE Loss): Calculate the initial and final kinetic energies.
$K E_{i} = \frac{1}{2} m v^{2}$
$K E_{f} = \frac{1}{2} (m + M) v_{f}^{2} = \frac{1}{2} (m + M) (\frac{m}{m + M} v)^{2} = \frac{1}{2} \frac{m ^{2}}{m + M} v^{2}$
The change in kinetic energy is $Δ K E = K E_{f} - K E_{i}$ , which is negative.
Thermodynamics (Energy Conversion): The increase in internal energy of the bullet-block system is equal to the magnitude of the kinetic energy loss.
$Δ U = - Δ K E = K E_{i} - K E_{f} = \frac{1}{2} m v^{2} - \frac{1}{2} \frac{m ^{2}}{m + M} v^{2} = \frac{1}{2} m v^{2} (1 - \frac{m}{m + M}) = \frac{1}{2} \frac{m M}{m + M} v^{2}$
This quantity, $\frac{1}{2} μ v^{2}$ where $μ$ is the reduced mass, represents the kinetic energy of relative motion that was converted to internal energy. If you are given the specific heat capacities, you can then calculate the final temperature.

Test your understanding!

A 10g lead bullet traveling at 500 m/s strikes a 1kg stationary block of wood and embeds itself. The specific heat of lead is 128 J/kg°C and that of wood is 1700 J/kg°C. Assuming the energy is distributed evenly, calculate the temperature rise of the block and bullet.

Show answer

Find the "lost" kinetic energy ( $Δ U$ ):
$m = 0.01$ kg, $M = 1$ kg, $v = 500$ m/s.
$Δ U = \frac{1}{2} \frac{m M}{m + M} v^{2} = \frac{1}{2} \frac{( 0.01 ) ( 1 )}{0.01 + 1} (500)^{2}$
$Δ U = \frac{1}{2} (\frac{0.01}{1.01}) (250000) \approx 0.5 \times (0.0099) \times 250000 \approx 1237.6 J$
Apply Calorimetry: This energy increases the temperature of the combined system.
$Δ U = (m_{b u l l e t} c_{b u l l e t} + m_{w oo d} c_{w oo d}) Δ T$
$1237.6 = ((0.01) (128) + (1) (1700)) Δ T$
$1237.6 = (1.28 + 1700) Δ T = 1701.28Δ T$
$Δ T = \frac{1237.6}{1701.28} \approx 0.7 3^{\circ} C$
The temperature of the system rises by approximately 0.73°C.

3. An Integrated Problem from a Different Perspective

Let's revisit a problem type that involves both work and heat, but analyze it from the perspective of energy conversion. The following JEE preparation material includes a problem where water is converted to steam. This is a process where heat supplied is converted into both internal energy and mechanical work of expansion.

Thermodynamics: Definition, Process, Equations & Laws

This resource from Allen Career Institute contains several solved thermodynamics problems. We will focus on one that illustrates the partitioning of energy according to the First Law.

Read question Q-3 in the 'Sample Questions On Thermodynamics' section. The problem asks for the change in internal energy when 1 kg of water is vaporized. Notice the steps: Heat supplied (Q) is given as the latent heat of vaporization. The change in volume (\Delta V) is calculated. The work done by the system during expansion is calculated as W = P\Delta V. The First Law, \Delta U = Q - W, is used to find that the heat supplied does two things: increases internal energy AND does work on the surroundings.

Now, consider the reverse process: 1 kg of steam at 373 K is compressed by a constant external pressure of 1 atm until it fully condenses into water at 373 K.

In this scenario:

Work is done on the system to compress it ( $W < 0$ ). This is macroscopic mechanical work being put into the system.
Heat is released from the system ( $Q < 0$ ), equal to the latent heat of condensation.
The change in internal energy, $Δ U = Q - W$ , will be a large negative number, as the steam (gas) has much higher internal energy than the water (liquid) at the same temperature.

This example shows how the First Law of Thermodynamics is a complete accounting system for energy. Mechanical work can be converted to internal energy, but energy can also be transferred as heat simultaneously.

Conclusion

Today we explored the fundamental principle that macroscopic mechanical work and microscopic internal energy are interconvertible, a concept encapsulated by the First Law of Thermodynamics.

Key Takeaways:

Equivalence of Work and Heat: Mechanical work done on a system and heat added to it are two ways of increasing its internal energy.
Adiabatic Work: In an adiabatic process ( $Q = 0$ ), any work done on a system ( $W < 0$ ) is fully converted into internal energy ( $Δ U = - W > 0$ ).
Dissipative Processes: Friction and inelastic collisions are prime examples where ordered kinetic energy is irreversibly converted into disordered internal energy, causing a temperature increase.
Energy Accounting: The First Law, $Δ U = Q - W$ , is the ultimate tool for tracking energy. To solve complex problems, you must correctly identify the system, the work done on/by it, and the heat transferred in/out of it.

Preview of the Next Lesson:
We are nearing the end of our journey through thermodynamics. In the next lesson, we will learn a very useful mathematical tool for your JEE Advanced preparation: how to apply the binomial approximation in thermal expansion and thermodynamic calculations for small changes. This technique is a powerful shortcut for simplifying complex expressions and is essential for solving certain problems quickly and accurately.

Recommended resources