Heat and Thermodynamics

Heat and Thermodynamics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Heat is energy in transit (Q, measured in joules); temperature (T, in kelvin) is a state property of matter, not a form of energy. The cornerstone formula is the First Law: ΔU = Q − W, where ΔU is the change in internal energy, Q is heat supplied to the system, and W is work done by the gas. For heating without phase change use Q = mcΔT (c = specific heat capacity, J kg⁻¹ K⁻¹); for phase change use Q = mL (L = latent heat, J kg⁻¹). The ideal gas law PV = nRT links pressure, volume and temperature. High-yield MDCAT pointers: distinguish isothermal (PV = const) from adiabatic (PV^γ = const); remember W = 0 when volume is fixed; entropy of an isolated system never decreases.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Core Concepts and the Laws

The Zeroth Law defines temperature: two systems each in thermal equilibrium with a third are in equilibrium with each other. The First Law is energy conservation applied to a gas: ΔU = Q − W (chemistry/biology sign convention, used in MDCAT). An adiabatic process has Q = 0, so ΔU = −W and temperature changes; an isothermal process has ΔU = 0 for an ideal gas, so Q = W. The Second Law states that entropy of an isolated system never decreases, which is why heat flows spontaneously from hot to cold and never the reverse without external work.

Work, Heat and Internal Energy

Work done by an expanding gas is W = PΔV for constant pressure, or the area under the curve on a P–V diagram in general. The four standard processes to recognise on a graph are:

• Isobaric (constant P): W = PΔV, Q = nCpΔT.
• Isochoric (constant V): W = 0, Q = nCvΔT, ΔU = Q.
• Isothermal: T constant, PV = const, ΔU = 0, Q = W.
• Adiabatic: Q = 0, PV^γ = const, TV^(γ−1) = const, where γ = Cp/Cv.

For an ideal gas, internal energy depends only on temperature: ΔU = nCvΔT.

Specific Heat and Latent Heat

Q = mcΔT governs sensible heating (raising temperature). The latent heats are different for different transitions: latent heat of fusion (Lf) for melting (water: 3.34 × 10⁵ J kg⁻¹) and latent heat of vaporisation (Lv) for boiling (water: 2.26 × 10⁶ J kg⁻¹). Thermal conductivity is governed by Q/t = kAΔT / L, where k (W m⁻¹ K⁻¹) is the material constant — high k for metals, low k for insulators.

Typical MDCAT Question Patterns

MCQs usually test (1) numerical Q = mcΔT problems with mass in grams and temperature in °C, (2) identifying which process has W = 0 or ΔU = 0, (3) the ratio γ = Cp/Cv (≈1.67 for monatomic, ≈1.4 for diatomic gases), and (4) direction of heat flow under the Second Law.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Subtleties

A frequent trap is sign convention. The MDCAT follows ΔU = Q − W (W = work done by the system). If a question uses ΔU = Q + W, W is work done on the system — re-derive carefully before plugging numbers. Another trap: adiabatic ≠ isothermal. In isothermal compression, heat must leave the gas to keep T constant; in adiabatic compression, no heat leaves, so T rises and PV^γ (not PV) stays constant.

For real gases, PV = nRT fails at high pressure or low temperature near condensation. The van der Waals equation corrects this, but MDCAT questions are restricted to ideal gases unless explicitly stated. Note also that Cp − Cv = R (per mole, J mol⁻¹ K⁻¹), so knowing one specific heat gives the other if γ is known.

Worked Example

A 0.040 kg copper calorimeter (c = 400 J kg⁻¹ K⁻¹) contains 0.200 kg of water (c = 4200 J kg⁻¹ K⁻¹) at 20 °C. A 0.050 kg iron block (c = 450 J kg⁻¹ K⁻¹) is heated to 100 °C and dropped in. Heat lost by iron = heat gained by water + calorimeter: 0.050 × 450 × (100 − T) = (0.200 × 4200 + 0.040 × 400) × (T − 20) 22500 − 225T = 856(T − 20) → 22500 − 225T = 856T − 17120 → T ≈ 39.6 °C. Check that no phase change applies, then report.

Practice Prompts

1. An ideal gas expands isothermally from 2 L to 4 L at 300 K against a constant external pressure of 1.5 atm. Find W, Q and ΔU. (Answer: W = PΔV ≈ 304 J, ΔU = 0, Q = 304 J.)
2. A refrigerator extracts 200 J of heat from a cold reservoir while 50 J of work is input. Is this consistent with the Second Law? Compute COP = Q_cold / W = 4; any value > 0 is permitted — yes, consistent.

Common Mistakes

• Confusing heat with temperature (energy in transit vs state property).
• Using Lf when Lv is required, or vice versa.
• Forgetting W = 0 in an isochoric process.
• Treating γ = 1.67 for all gases — diatomic gases at room temperature use γ ≈ 1.4.

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