Why Leaving the Fridge Door Open Heats Your Kitchen

Mechanics follows how a single ball moves. Thermodynamics looks at a gas of countless molecules as a whole, using a few measurable quantities — pressure, volume, temperature — to track how energy flows in as heat and out as work.

This part covers thermal equilibrium and the zeroth law, heat, work and internal energy, the first law, and first-law problems. Take $R=8.314$ J/mol K.

Two bodies in contact are in thermal equilibrium when no net heat flows between them; the zeroth law states that if A and B are each in thermal equilibrium with C, then A and B are in thermal equilibrium with each other — and the property they share is temperature.

Worked example — reaching equilibrium. In an insulated container, $2$ kg of water at $60$ °C is mixed with $3$ kg of water at $20$ °C. At equilibrium both share one temperature $T$:

$$2(60-T)=3(T-20)\Rightarrow 120+60=5T\Rightarrow T=36\text{ °C}$$

An everyday example. A doctor keeps a thermometer under your tongue for a minute so that it comes into thermal equilibrium with your body before it is read.

The substance. Without the zeroth law, a thermometer reading would say nothing about two bodies it never touched together — it is what makes temperature a meaningful quantity.

Heat is energy transferred because of a temperature difference, work is energy transferred by a force acting through a displacement, such as a moving piston, and internal energy is the total kinetic and potential energy of the molecules stored in the system.

- **Internal energy $U$ — a state variable: it depends only on the present state of the system
- Heat $Q$ and work $W$ — not state variables: they depend on the path taken between states

Worked example 1 — work by a gas.** A gas expands at a constant pressure of $1\times 10^5$ Pa from $1$ L to $3$ L:

$$W=P\Delta V=1\times 10^5\times (3-1)\times 10^{-3}=200\text{ J}$$

Worked example 2 — internal energy. For $1$ mol of an ideal monatomic gas at $300$ K:

$$U=\frac{3}{2}\mu RT=\frac{3}{2}\times 1\times 8.314\times 300\approx 3740\text{ J}$$

For an ideal gas, $U$ depends only on temperature.

An everyday example. Rubbing your palms together (work) and holding a hot cup of chai (heat) both warm your hands — two different transfers raising the same internal energy.

The substance. A body contains internal energy, never "heat" or "work" — those words describe energy only while it is crossing the boundary.

**The first law states that heat supplied to a system equals the increase in its internal energy plus the work done by the system: $Q=\Delta U+W$, or $\Delta U=Q-W$.

Sign conventions used here:

- $Q>0$ when heat flows into** the system; $Q<0$ when it leaves
- **$W>0$ when the system does work on the surroundings** (expansion); $W<0$ when work is done on it (compression)
- **$\Delta U>0$ when internal energy rises

Worked example 1.** A gas absorbs $500$ J of heat and does $200$ J of work while expanding:

$$\Delta U=500-200=300\text{ J}$$

Worked example 2. A gas is compressed with $150$ J of work done on it while it gives out $100$ J of heat:

$$\Delta U=Q-W=(-100)-(-150)=+50\text{ J}$$

Its internal energy rises, so it warms up.

An everyday example. The barrel of a bicycle pump gets warm as you pump, because work done on the air raises its internal energy faster than heat can escape.

The substance. **Some books write $\Delta U=Q+W$ with $W$ as work done on the system** — the physics is the same, so always state the convention you use.

**Identify the process, find the two quantities you can calculate directly — often $W$ from $P\Delta V$ and $\Delta U$ from the temperature change — and use $Q=\Delta U+W$ for the third.

Worked example 1 — constant pressure.** $2$ mol of an ideal monatomic gas at $1\times 10^5$ Pa is heated from $300$ K to $400$ K.

$$W=P\Delta V=\mu R\Delta T=2\times 8.314\times 100\approx 1663\text{ J}$$

$$\Delta U=\frac{3}{2}\mu R\Delta T\approx 2494\text{ J}Q=\Delta U+W\approx 4157\text{ J}$$

Worked example 2 — boiling. $1$ kg of water at $100$ °C becomes steam at $1.013\times 10^5$ Pa, its volume growing from $1.0\times 10^{-3}$ m${}^3$ to $1.671$ m${}^3$:

$$W=1.013\times 10^5\times 1.670\approx 1.69\times 10^5\text{ J}\Delta U=2.26\times 10^6-1.69\times 10^5\approx 2.09\times 10^6\text{ J}$$

An everyday example. Boiling water in an open pan, most of the heat goes into pulling molecules apart; only a small part pushes back the atmosphere.

The substance. **$\Delta U$ depends only on the start and end states**, so any path between the same states gives the same $\Delta U$.