Heat and Thermodynamics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Heat (Q) is energy in transit, measured in joules (J). Temperature (T/θ) is a state variable measured in kelvin (K) or Celsius (°C), with T(K) = θ(°C) + 273.
Specific heat capacity: Q = mcΔθ, where m is mass (kg), c is specific heat capacity (J kg⁻¹ K⁻¹), Δθ is temperature change. Water has c = 4200 J kg⁻¹ K⁻¹.
Latent heat (phase change at constant T): Q = mL, where L is specific latent heat (J kg⁻¹). Use this — not mcΔθ — when ice melts or water boils.
Ideal gas law: PV = nRT, with R = 8.314 J mol⁻¹ K⁻¹. Temperature must be in kelvin.
First Law: ΔU = Q − W (W = work done by the gas = pΔV).
Second Law: heat flows spontaneously from hot to cold; entropy of an isolated system never decreases.
Three heat-transfer modes: conduction, convection, radiation.
Thermal expansion relations: β = 2α (areal), γ = 3α (cubical), where α is linear expansivity (K⁻¹).

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Temperature, Heat, and Internal Energy

Temperature measures the average kinetic energy of molecules; heat is energy flowing due to a temperature difference. Internal energy (U) is the total microscopic kinetic + potential energy of a system. For an ideal gas, U depends only on temperature, so ΔU = 0 during an isothermal process.

Heat Capacity vs Latent Heat

Heating a substance raises its temperature until it begins changing phase. During the phase change, temperature stays constant while energy is absorbed as latent heat:

Quantity	Formula	When Used
Sensible heat	Q = mcΔθ	Temperature changing, no phase change
Latent heat	Q = mL	Melting, boiling, freezing, condensation
Gas work	W = pΔV	Gas expanding/compressing against pressure

The Gas Laws and Ideal Gas Equation

Boyle’s Law: PV = constant (constant T)
Charles’s Law: V/T = constant (constant P)
Pressure Law: P/T = constant (constant V)
Combined: PV = nRT (always use T in kelvin)

For an isothermal change, W = nRT ln(V₂/V₁); for an isobaric change, W = pΔV.

Thermal Expansion

Solids expand when heated. Linear: ΔL = αL₀Δθ. Areal: ΔA = 2αA₀Δθ. Cubical: ΔV = γV₀Δθ, with γ = 3α. This is why bridges have expansion joints and bimetallic strips bend on heating.

First and Second Laws

First Law (conservation of energy): ΔU = Q − W. W is work done by the system. Second Law: heat naturally passes from hot to cold; entropy (S) of an isolated system increases (ΔS ≥ 0). Heat engines cannot be 100% efficient — efficiency η = 1 − T_cold/T_hot (Kelvin).

Typical JAMB Question Patterns

Numerical on Q = mcΔθ or Q = mL (e.g., energy to melt a block of ice).
PV = nRT calculation, often testing the kelvin conversion trap.
Identifying isothermal/adiabatic/isobaric/isochoric lines on a PV diagram.
Sign convention in the first law: is heat added or removed? Does the gas do work or have work done on it?

Tip: Always convert °C to K before plugging into any gas law. Missing this single step costs JAMB candidates 1–2 marks every year.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Process-by-Process Comparison

Process	Held constant	First-law form	Work done
Isothermal	T	Q = W (since ΔU = 0)	W = nRT ln(V₂/V₁)
Isobaric	P	Q = ΔU + pΔV	W = pΔV
Isochoric	V	Q = ΔU (W = 0)	0
Adiabatic	Q = 0	ΔU = −W	W = ½(p₁V₁ − p₂V₂)

On a PV diagram, isothermal curves are gentler (less steep) than adiabats; adiabats rise more sharply because no heat escapes.

Edge Cases and Examiner Traps

Sign convention mismatch — writing ΔU = Q + W. The convention in JAMB’s Physics syllabus has W as work done by the gas, so ΔU = Q − W.
Phase-change oversight — using Q = mcΔθ across melting/boiling points. Latent heat is absorbed without temperature change.
Linear vs cubical confusion — γ ≈ 3α is exact for isotropic solids but only an approximation; for crystals with direction-dependent expansion, separate coefficients apply.
Specific vs molar — specific heat capacity c (J kg⁻¹ K⁻¹) differs from molar heat capacity Cₘ (J mol⁻¹ K⁻¹). For monatomic ideal gases, Cᵥ = 3R/2; for diatomic, Cᵥ = 5R/2.
Reversible vs irreversible — JAMB may distinguish heat flow through a perfect conductor (reversible idealisation) from real-world spontaneous heat flow.

Connections to Other Topics

Kinetic theory links PV = nRT to microscopic motion: PV = ⅓ Nm⟨c²⟩, giving the average kinetic energy per molecule as 3kT/2, where k = R/Nₐ = 1.38 × 10⁻²³ J K⁻¹. This same equipartition result underlies specific heat capacities of gases. Thermal conductivity (Fourier’s law) is tested in the same chapter as conduction; expect numericals on heat flow through composite walls.

Worked Micro-Example

Question: 0.02 kg of ice at 0 °C is melted and then heated to 30 °C. Find the total heat required. (L_f = 3.34 × 10⁵ J kg⁻¹, c_water = 4200 J kg⁻¹ K⁻¹)

Step 1 — melting: Q₁ = mL = 0.02 × 3.34 × 10⁵ = 6680 J
Step 2 — heating water: Q₂ = mcΔθ = 0.02 × 4200 × 30 = 2520 J
Total Q = 6680 + 2520 = 9200 J

Practice Prompts

A gas in a cylinder expands isothermally from 0.004 m³ to 0.010 m³ at 300 K against a constant external pressure of 1.0 × 10⁵ Pa. Find the work done and the heat absorbed. (Hint: ideal gas → use nRT relation or pΔV carefully.)
A brass rod of length 2.00 m is heated from 20 °C to 120 °C. Given α_brass = 1.9 × 10⁻⁵ K⁻¹, calculate the new length and the percentage change. (Watch for consistent units in Δθ.)

Strategy for JAMB: Allocate about 60–90 seconds per Heat & Thermodynamics question. The 3–4 questions per sitting are almost always gas-law numericals plus one conceptual first/second-law item. Master the kelvin conversion and the sign convention — together they remove the two most common marks lost on this topic.

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Heat and Thermodynamics