Physics › 🔥 Thermodynamics

The Laws of Thermodynamics

Why you can spend energy, but you can never spend all of it twice.

AP Physics 2 levelUni Year 1
💡
The big idea: Energy is never created or destroyed — the first law makes that a hard accounting rule. But the second law adds a twist: every real process leaks a little energy into a form nobody can use for work again. That one-way slide toward disorder is why heat engines always waste some heat, and why time itself seems to have a direction.
🎯 By the end, you'll be able to
  • Apply the first law, ΔU = Q − W, to find unknown heat, work, or internal-energy changes in a thermodynamic process.
  • Explain what entropy actually measures and why the second law makes some energy permanently unusable for work.
  • Calculate the maximum (Carnot) efficiency of a heat engine from its hot and cold reservoir temperatures.
  • Spot and avoid the classic sign-convention trap that flips ΔU = Q − W into ΔU = Q + W.
📎 You should already know
  • Ideal Gas Law & Kinetic Theory
  • Work Done by a Gas (PV Diagrams)
  • Temperature, Heat, and Thermal Equilibrium

Energy's Bookkeeping — and Its One-Way Street

Thermodynamics starts with a deceptively simple question: when you heat something up, push on it, or let it expand, where does the energy actually go? The first law answers that with pure accounting — energy in, energy out, nothing lost. But then the second law shows up and complicates the tidy picture: some of that energy, once spent, can never be fully recovered. That one-way street is why your coffee cools down but never spontaneously reheats itself, and why every engine ever built — no matter how cleverly designed — throws away some heat as waste.

🔑 The First Law: Energy Is Conserved, Full Stop

Whatever energy enters a system as heat, minus whatever leaves as work done by the system, is exactly how much the system's internal energy changes. No energy is created or destroyed — it's just moved around and converted between forms.

\[ \Delta U = Q - W \]
ΔU = change in internal energy, Q = heat added to the system, W = work done by the system on its surroundings (positive when the gas expands).
\[ W = P \Delta V \]
For a gas expanding or compressing at constant pressure, the work it does equals pressure times the change in volume.
\[ \eta_{max} = 1 - \dfrac{T_C}{T_H} \]
The Carnot efficiency — the theoretical ceiling for any heat engine operating between a hot reservoir at T_H and a cold reservoir at T_C (both in kelvin).
🎮 Interactive: Piston, Heat, and Work in Action LIVE
Push the piston, add heat, and watch ΔU, Q, and W update in real time — a hands-on feel for the first law.

The Second Law: Why Energy Degrades

The second law says that in any real process, the total entropy of an isolated system never decreases — it stays constant only in an idealized reversible process, and increases in every real one. Entropy is often described as "disorder," but a more useful intuition is that it measures how spread out and unrecoverable energy has become. High-grade energy, like a compressed spring or a hot gas, can do useful work. Once that energy disperses into random molecular motion — heat spread evenly through a room, say — you can't gather it all back up. That's also why a heat engine can never turn 100% of the heat it absorbs into work: some heat must always be dumped into a colder reservoir just to keep the entropy books balanced.

📝 Worked example: A gas in a cylinder absorbs 500 J of heat from a flame. As it expands, it pushes the piston outward, doing 200 J of work on its surroundings. What is the change in the gas's internal energy?
  1. Identify the given quantities using the convention Q > 0 for heat added and W > 0 for work done by the gas: Q = +500 J, W = +200 J.
  2. Apply the first law: \(\Delta U = Q - W\).
  3. Substitute: \(\Delta U = 500\,\text{J} - 200\,\text{J} = 300\,\text{J}\).
✓ ΔU = +300 J — the gas's internal energy rises by 300 J, since it absorbed more heat than it gave up as work.
📝 Worked example: A heat engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K, absorbing 2000 J of heat from the hot reservoir each cycle. Find the engine's maximum possible efficiency and the maximum work it can output per cycle.
  1. Apply the Carnot efficiency formula: \(\eta_{max} = 1 - \dfrac{T_C}{T_H}\).
  2. Substitute the reservoir temperatures (already in kelvin): \(\eta_{max} = 1 - \dfrac{300}{600} = 1 - 0.5 = 0.5\), or 50%.
  3. Find the maximum work using \(W_{max} = \eta_{max} \times Q_H\): \(W_{max} = 0.5 \times 2000\,\text{J} = 1000\,\text{J}\).
✓ Maximum efficiency = 50%, maximum work output = 1000 J per cycle. The remaining 1000 J is exhausted as waste heat to the cold reservoir — not lost, just no longer usable for work.
⚠️ Common Trap: Whose Work Is It, and Why No Engine Hits 100%

Watch the sign convention carefully — this lesson uses \(W\) as work done by the system (common in physics courses), but many chemistry courses define \(W\) as work done on the system, flipping the equation to \(\Delta U = Q + W\). Both describe the same physics; only the bookkeeping differs, so always check which convention a problem is using before you plug in numbers.

And no matter how well-engineered a heat engine is — even one with zero friction and perfect insulation — the second law ensures it can never convert 100% of absorbed heat into work. Some heat must always be rejected to a cooler reservoir. That's not an engineering flaw; it's a hard boundary set by the physics itself.

Check your understanding

1. In the equation ΔU = Q − W, as used in this lesson, what does W represent?
In this convention, W is the work done by the system (the gas) on its surroundings — that's why positive W drains internal energy unless offset by heat input.
2. A gas releases 150 J of heat to its surroundings while the surroundings do 300 J of work on the gas (compressing it). What is the change in the gas's internal energy?
Heat leaving the gas makes Q = −150 J. Work done ON the gas means W (work done by the gas) is −300 J. So ΔU = Q − W = −150 − (−300) = +150 J: the compression adds more energy than the heat loss removes.
3. Why can't a heat engine achieve 100% efficiency, even with zero friction and perfect insulation?
The second law forbids a cyclic heat engine from converting 100% of absorbed heat into work — some heat must always flow to a colder reservoir, no matter how ideal the engine's construction.
4. A power plant's furnace runs at 800 K and rejects waste heat to a river at 300 K. What is the maximum possible (Carnot) efficiency?
η_max = 1 − T_C/T_H = 1 − 300/800 = 1 − 0.375 = 0.625, or 62.5%. Real plants achieve less due to friction and irreversibilities, but this is the theoretical ceiling.
✅ Key takeaways
  • The first law, ΔU = Q − W, is just energy bookkeeping: what goes in as heat, minus what leaves as work, is what's left over as internal energy.
  • The second law adds a one-way arrow: entropy in an isolated system never decreases, which is why heat flows from hot to cold and why energy quality degrades over time.
  • Every heat engine must reject some heat to a cold reservoir — the Carnot efficiency η = 1 − T_C/T_H is the absolute best any engine can do between two given temperatures.
  • Sign conventions vary between textbooks (physics vs. chemistry) — always confirm whether W means work done by or on the system before solving a problem.