Lesson 14: Thermodynamics — Heat, Temperature, and the Laws

Thermodynamics is the physics of heat and its transformations — and its laws are the most unforgiving in science. Engines, refrigerators, power stations and the climate all obey them. This lesson covers specific heat capacity, the first law (energy conservation applied to gases), and efficiency bounded by the second law.

Heat, temperature and the first law

Temperature measures the average kinetic energy of particles; heat is energy *in transit* between temperatures. Heating a material without a change of state follows Q = mcΔT, where c is the specific heat capacity — for water 4,186 J/(kg °C), one of the highest of common materials, which is why water stores and transports heat so well and why coastal climates are mild.

The first law of thermodynamics is energy conservation for a gas: ΔU = Q − W, where ΔU is the change in internal energy (the total kinetic and potential energy of the particles), Q is heat added *to* the gas, and W is work done *by* the gas. For an ideal gas, internal energy depends only on temperature. A gas expanding against a piston at constant pressure does work W = PΔV. Compress a bicycle-pump quickly and it warms: work done *on* the gas raises ΔU with no time for heat to escape.

The second law forbids perfection: no heat engine can convert all absorbed heat into work; some must be rejected to a colder reservoir. Efficiency e = W ÷ Q_hot = 1 − Q_cold ÷ Q_hot, and the theoretical maximum (Carnot) efficiency is 1 − T_cold ÷ T_hot with temperatures in kelvin. Entropy — the universe's bookkeeping of disorder — never decreases overall.

Worked example 1: heating the kettle

How long does a 2,000 W kettle, 85% efficient, take to heat 0.50 kg of water from 20 °C to 80 °C? Q needed = mcΔT = 0.50 × 4,186 × 60 = 125,600 J ≈ 1.26 × 10⁵ J. Useful power = 0.85 × 2,000 = 1,700 W. t = Q ÷ P = 125,600 ÷ 1,700 = 74 s. (Real kettles lose heat to the jug and the air, and switch off after boiling begins — measurements of 80–90 s would be typical.)

Worked example 2: applying the first law

A gas absorbs 500 J of heat while expanding and doing 180 J of work on a piston. Its internal energy change is ΔU = Q − W = 500 − 180 = +320 J, so its temperature rises (for an ideal gas). Check a work term: expansion at constant pressure 1.2 × 10⁵ Pa with ΔV = 5.0 × 10⁻⁴ m³ gives W = PΔV = 1.2 × 10⁵ × 5.0 × 10⁻⁴ = 60 J. Insulating the cylinder (Q = 0) during such an expansion would give ΔU = −60 J — adiabatic expansion cools the gas, the principle behind the chill on a deodorant can.

In the lab: specific heat of aluminium

An electric immersion heater embedded in a 0.20 kg aluminium block runs at 12 V and 3.5 A for 300 s. The block's temperature rises 58 °C. Electrical energy supplied = VIt = 12 × 3.5 × 300 = 12,600 J. Apparent c = Q ÷ (mΔT) = 12,600 ÷ (0.20 × 58) = 1,086 J/(kg °C) — but the accepted value is 900. The method overestimates c by about 20% because some of the electrical energy escaped to the bench, air and thermometer instead of staying in the block. Lagging the block with foam, stirring-equivalent design (the heater hole is snug, a drop of oil improves contact), and heating for a shorter interval in a pre-warmed room all shrink the error. The key lesson: a systematic loss always biases the result in a *predictable direction* — here upwards, because we credited the block with energy it never received.

Physics in the real world

Power stations are heat engines limited by steam temperatures: even a modern plant at 600 °C rejects enough heat to run at roughly 40–45% efficiency, a Carnot ceiling of about 65% at those temperatures. Refrigerators and heat pumps move heat "uphill" using work; a heat pump delivering 3 kW of heating for 1 kW of electricity is not breaking the efficiency rules — its coefficient of performance (heat moved ÷ work input) can legitimately exceed 1, because it relocates heat rather than converting work into it. Home insulation, double glazing and heat-recovery ventilation are applied first-law accounting.

Common mistakes

• Saying an object "contains heat" — objects contain internal energy; heat is transfer.
• Believing "cold" flows into a room; only heat flows, from hot to cold.
• Sign errors in ΔU = Q − W: fix the convention (Q in, W done by gas) and stick to it.
• Computing Carnot efficiency with Celsius temperatures — absolute kelvin is mandatory.
• Treating a heat pump's COP of 3 as "300% efficiency" and claiming perpetual motion; the missing energy comes from the outside air.

Exam Experience

1. The temperature of a substance is a measure of: A) the total internal energy it contains. B) the average kinetic energy of its particles. C) the heat it can release. D) its specific heat capacity. Answer: B. Temperature reflects the *average* particle kinetic energy; total internal energy also depends on mass and particle potential energy.

2. A gas does 250 J of work while its internal energy increases by 150 J. The heat added to the gas was: A) 100 J. B) 250 J. C) 400 J. D) −100 J. Answer: C. ΔU = Q − W, so Q = ΔU + W = 150 + 250 = 400 J.

3. A heat engine absorbs 1,000 J and rejects 700 J each cycle. Its efficiency is: A) 70%. B) 43%. C) 30%. D) 15%. Answer: C. Work = 1,000 − 700 = 300 J; efficiency = 300 ÷ 1,000 = 30%.

Constructed response (5 marks). A heat engine absorbs 1,200 J of heat from a hot reservoir each cycle and rejects 850 J to a cold reservoir. The reservoirs are at 600 K and 300 K. (a) Calculate the work done per cycle. (b) Calculate the engine's efficiency. (c) Calculate the maximum possible (Carnot) efficiency between these temperatures and comment on the comparison. Scoring guide: (a) 1 mark for W = 1,200 − 850 = 350 J. (b) 1 mark for e = 350 ÷ 1,200 = 29% (0.29). (c) 1 mark for using kelvin (already given) in e_max = 1 − 300 ÷ 600; 1 mark for 50%; 1 mark for a comment that the actual engine falls short of the Carnot limit because of friction, heat losses and non-ideal (non-reversible) processes — but can never exceed it. Common error: using 600 °C and 300 °C style values or computing efficiency as W ÷ Q_cold.

Going further: changes of state and latent heat

Heating does not always mean warming. During melting or boiling, added energy breaks intermolecular bonds without raising the temperature at all — it appears as latent heat. Ice's latent heat of fusion is 3.34 × 10⁵ J/kg; water's latent heat of vaporisation is 2.26 × 10⁶ J/kg. Worked example: turn 0.50 kg of ice at 0 °C into water at 20 °C. Melting: Q = mL_f = 0.50 × 3.34 × 10⁵ = 167,000 J. Warming the meltwater: Q = mcΔT = 0.50 × 4,186 × 20 = 41,860 J. Total = 209,000 J ≈ 2.1 × 10⁵ J — four-fifths of it spent on the change of state, not the temperature rise. This is why ice is such an effective coolant: it absorbs enormous energy while staying at 0 °C. Latent heat runs the weather. Evaporating 1 kg of ocean water stores 2.26 × 10⁶ J; when that vapour condenses inside a hurricane, the energy returns as heat that powers the storm's winds. Sweating cools you by exactly the same physics in reverse: each gram evaporated carries away about 2.4 kJ of body heat. Steam burns are worse than boiling-water burns because condensing steam dumps its latent heat into the skin on top of its temperature — a painful lesson in phase-change accounting.

In focus: absolute zero and the kelvin bargain

Efficiency calculations demand kelvin because the Celsius zero is arbitrary — an engine working between 600 °C and 300 °C does *not* have a Carnot efficiency of 50%: converted properly (873 K and 573 K), it is 1 − 573 ÷ 873 = 34%. Only ratios of absolute temperatures are physically meaningful, since thermal energy content scales from absolute zero. Kelvin also prices the internal energy of a gas: for an ideal gas, the average translational kinetic energy per mole is (3/2)RT; at 300 K that is 1.5 × 8.31 × 300 = 3,740 J per mole. Heating a mole of gas by 10 K at constant volume takes only about 125 J of that store — the small change that pressure and thermometer readings amplify into measurables. Approach absolute zero and the third law takes over: it cannot be reached in any finite number of steps, though laser-cooled atoms have been taken below a billionth of a kelvin. Superconductivity and superfluidity live down there — materials properties that vanish entirely at everyday temperatures — making the cold end of the scale as physically rich as the stellar end. Temperature, this course's quietest quantity, spans from 10⁻¹⁰ K in atomic physics laboratories to 10⁷ K in the Sun's core: seventeen orders of magnitude governed by one idea, average kinetic energy.

Formula card

Exam technique: declare your sign convention in the first line and never renegotiate it mid-question. For Carnot questions, convert to kelvin before doing anything else — examiners set the Celsius trap deliberately, and it catches someone in every cohort.