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Playbook

Heat and Thermodynamics

39 q · 21% HARD. Calorimetry mixing (heat-balance equations with phase change) + thermodynamic process variants (PV=nRT, PVⁿ=const, P=kT) are the HARD pools. Temperature-scale conversions and latent-heat statement-truth are the EASY plumbing.

questions in the bank
39
tagged HARD
21%
subtopic(s)
4
worked examples
2

When you’ll see it

A Q = mcΔT calorimetry mixing, a temperature-scale conversion, a phase-change with latent heat, a PV = nRT or PVⁿ = const process variant.

How this chapter is tested

39 q in 10 years · 21% HARD. Four subtopics: Heat / Calorimetry / Specific Heat (13 q · 31% HARD), Temperature and Thermometry (11 q), Phase Change and Boiling (9 q), Thermodynamic Processes (6 q · 33% HARD).

Calorimetry is the marquee. Q_gained_by_cold = Q_lost_by_hot (no external heat transfer). Sensible heat Q = mcΔT, latent heat Q = mL at phase boundaries. The 'ice at −10°C + water at 30°C, find final T' problem has THREE segments: warm ice −10→0 (sensible), melt ice at 0 (latent), warm melted-ice water 0→T (sensible). Cool original water 30→T (sensible). Set up the heat balance: m_ice·c_ice·10 + m_ice·L_f + m_ice·c_w·T = m_w·c_w·(30−T).

Thermodynamic processes are tested as algebra. PV = nRT is the ideal gas law. Process variants: isothermal PV=const (Boyle); isobaric V/T=const; isochoric P/T=const; adiabatic PVᵞ=const. NDA HARD shape: 'in a process PV² = const for ideal gas, find T₁/T₂ vs V₁/V₂' — derive from PV=nRT plus PV² = const: T·V = const ⟹ T₁/T₂ = V₂/V₁.

The sub-skills

The rules and habits that decide whether you get a question right.

  • Heat balance equation

    ∑Q_gained = ∑Q_lost. Each substance contributes mcΔT for sensible heating, mL for any phase boundary crossed. Final state (all liquid? mixed?) is part of the setup, not the output.

  • Temperature scale conversion

    T_C = (T_F − 32) × 5/9. T_K = T_C + 273.15. T_F = T_C × 9/5 + 32. Scales cross: C=F at −40°. C=K never (offset 273). K=F at 574.25.

  • PV = nRT process variants

    Isothermal: T const, PV=const. Isobaric: P const, V/T=const. Isochoric: V const, P/T=const. Adiabatic: PVᵞ=const, T·V^(γ−1)=const, T·P^((1−γ)/γ)=const.

  • Latent heat for phase changes

    Water: L_f ≈ 334 kJ/kg (melt at 0°C). L_v ≈ 2260 kJ/kg (boil at 100°C). Phase change happens AT temperature, no T change while changing phase. Q = mL.

2 worked examples from the bank

Real past-year questions illustrating the playbook. Click to reveal options + solution.

Example 1Heat and ThermodynamicsHARD
A 5 g piece of ice at −20 °C-20\,°C is put into mm kg of water at 30 °C30\,°C. The final temperature is 0 °C0\,°C in liquid phase. What is the value of mm in kg?

[Q61 · Apr · 2026]

Example 2Heat and ThermodynamicsHARD
In a certain process, PV2=constantPV^2 = \text{constant} for an ideal gas. If initial temperature is T1T_1, final temperature T2T_2, initial volume V1V_1, final volume V2V_2, then which one is correct?

[Q54 · Apr · 2026]

Traps to expect

Distractor shapes specific to this chapter. The page-wide Traps section covers the bank-level patterns.

  • Skipping a latent-heat term

    Forgetting to add Q = mL when crossing the 0°C or 100°C boundary. The wrong option matches the no-latent calculation. ALWAYS map the temperature journey of EACH substance first.

  • Process variant misidentification

    Assuming PV = const (isothermal) when the question stipulates a different relationship. Always read the process specification carefully; derive from PV = nRT + the given constraint, don't fall back on a memorised pair.

  • Specific heat 'depends on mass and shape'

    Specific heat is intensive — depends only on material. The wrong option claims dependence on mass or shape, both wrong.

Drill every heat and thermodynamics question

39 questions from the bank, scoped to 4 bundled subtopics.

Drill the 39 questions

Related playbooks

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