Internal energy work and heat transfer

The internal energy $U$ of a body is the sum of the kinetic energy of the chaotic motion of its particles and the potential energy of their interaction. Because chaotic motion is thermal motion, $U$ grows with temperature: for a monatomic ideal gas $U = \frac{3}{2}\nu R T$. Internal energy can be changed in exactly two ways — by doing work and by heat transfer. The first law of thermodynamics is written as $\Delta U = Q + A_{\text{ext}},$ where $Q$ is the heat received by the body and $A_{\text{ext}}$ is the work done on it by external forces. If a gas itself expands and does work $A_{\text{gas}}$, then $A_{\text{ext}} = -A_{\text{gas}}$, and the law becomes $\Delta U = Q - A_{\text{gas}}$. The signs carry the physics: $Q > 0$ when the body absorbs heat and $A_{\text{ext}} > 0$ when external forces compress the body, while $Q < 0$ and $A_{\text{ext}} < 0$ mean heat is released and the body expands on its own.

A gas receives $Q = 300$ J of heat from a heater and, while expanding, does work $A_{\text{gas}} = 120$ J. Find the change in the internal energy of the gas.

Air in a cylinder is compressed: external forces do $200$ J of work on the gas, while the gas releases $150$ J of heat to the surroundings. Find the change in internal energy and state how the temperature changed.

A gas undergoes a closed cycle $1 \to 2 \to 3 \to 1$.

• $1 \to 2$: the gas absorbs $400$ J of heat and does $150$ J of work.
• $2 \to 3$: isochoric cooling, internal energy decreases by $250$ J, no work is done.
• $3 \to 1$: external forces do $100$ J of work on the gas.

Find the heat in segments $2 \to 3$ and $3 \to 1$, the work done by the gas over the whole cycle, and verify that the total $\Delta U$ for the cycle is zero.

Self-check in a simulation (safe: no heaters, no open flame).

1. Open the PhET «Gas Properties» simulation (or an equivalent) and choose an ideal monatomic gas in a closed container with a piston.
2. Record the initial values of $T$ and $U$ (or of volume and pressure).
3. Run A (heat transfer only): lock the piston and supply heat. Verify that $A_{\text{gas}} = 0$ and $\Delta U = Q$.
4. Run B (work only): remove the heat source and slowly compress the gas with the piston. Verify that $Q = 0$ and $\Delta U = A_{\text{ext}} > 0$.
5. Run C (both mechanisms): compress the gas while removing heat, and find $\Delta U = Q + A_{\text{ext}}$.
6. Record in a table for each run: $Q$, $A_{\text{ext}}$, the calculated $\Delta U$ and the observed change in $T$; mark the rows where the signs disagreed with your expectation.

Conclusion — 3 sentences: which two mechanisms changed $U$, how the sign of $Q$ matched the direction of heat flow, and what happened to $U$ during adiabatic compression.

Show that exactly two channels change internal energy — heat transfer and work — and fix the sign rule for $Q$ and $A_{\text{ext}}$ before numbers are substituted into the formula.

• Diagram: a rectangle labelled system (gas) inside a dashed frame labelled environment.
• Incoming arrows: $Q > 0$ (a heater supplies heat) and $A_{\text{ext}} > 0$ (the piston is pushed in).
• Outgoing arrows: $Q < 0$ (heat flows to a cooler) and $A_{\text{ext}} < 0$ (the expanding gas pushes the piston out).
• Beside it — a bar chart for three cases: $Q$ only, $A_{\text{ext}}$ only, and both together; the bar height equals $\Delta U$.
• On the right — a graph of $U$ versus $T$ for an ideal gas: a straight line through the origin, showing that $\Delta U > 0$ always means a temperature rise.

You inflate a bicycle tyre with a hand pump. On each stroke you push the piston with a force of about $30\ \mathrm{N}$ through $0{,}30\ \mathrm{m}$, doing work $A = F s = 9\ \mathrm{J}$. Over 50 strokes about $450\ \mathrm{J}$ goes into the air inside the cylinder, and the metal barrel becomes noticeably warm — several degrees hotter to the touch than the room air.

Internal energy changes not only through heat transfer but also when work is done. The compression is fast, so heat exchange with the surroundings has no time to occur ($Q \approx 0$); the first law of thermodynamics then gives $\Delta U = A$: the work of the external force goes into the kinetic energy of the air molecules and its temperature rises.

A $2200\ \mathrm{W}$ kettle heats $1{,}5\ \mathrm{L}$ of water in $4\ \mathrm{min}$. The energy delivered to the water is $Q = P t = 2200 \cdot 240 = 5{,}28 \cdot 10^{5}\ \mathrm{J}$. With water's specific heat capacity $4200\ \mathrm{J/(kg\cdot K)}$ the $1{,}5\ \mathrm{kg}$ mass warms by $\Delta T = \dfrac{5{,}28 \cdot 10^{5}}{1{,}5 \cdot 4200} \approx 84\ \mathrm{K}$ — from $20\ ^\circ\mathrm{C}$ almost to boiling.

Here nothing compresses or moves the water, so the mechanical work is zero and the change in internal energy is entirely due to heat transfer: $\Delta U = Q$. The heater's power sets the rate of that transfer, and the heat capacity links the received energy to the temperature rise.

A car of mass $1200\ \mathrm{kg}$ brakes from $25\ \mathrm{m/s}$ (that is $90\ \mathrm{km/h}$) to a stop. Its kinetic energy $E_k = \dfrac{mv^2}{2} = \dfrac{1200 \cdot 25^2}{2} = 3{,}75 \cdot 10^{5}\ \mathrm{J}$ turns into heat in the brake discs and tyres. If one cast-iron brake disc of mass $8\ \mathrm{kg}$ ($c \approx 460\ \mathrm{J/(kg\cdot K)}$) absorbed all of it, it would heat up by $\Delta T = \dfrac{3{,}75 \cdot 10^{5}}{8 \cdot 460} \approx 100\ \mathrm{K}$.

The braking friction force does negative work on the car, and precisely that work increases the internal energy of the disc and pads: mechanical energy does not vanish but is redistributed into the energy of chaotic molecular motion. The heated parts then give it back to the air, now by heat transfer.