Thermodynamics, Heat and Calorimetry (AAMC 5E)

Thermodynamics relates heat, work, internal energy, and temperature. For the MCAT Chemical and Physical Foundations section, know the temperature scales, the zeroth/first/second laws, calorimetry (q = mcΔT), latent heat of phase changes, work by a gas on PV diagrams, entropy, and the three modes of heat transfer. Numerical items are high-yield and strictly algebra-based (no calculus).

MCAT content mapping. Thermodynamics spans temperature & the zeroth law, heat transfer (conduction, convection, radiation), specific & latent heat, the first law (ΔU = Q − W), PV work and processes, the second law & entropy, and heat engines/efficiency.

Temperature & Its Scales

Temperature measures the average translational kinetic energy of the particles of a substance. Three scales appear on the MCAT:

Celsius (°C): water freezes at 0°C, boils at 100°C.
Kelvin (K): the absolute scale; 0 K = absolute zero = −273.15°C. Same size degree as Celsius: T(K) = T(°C) + 273.15. Always use kelvin in gas-law and efficiency formulas.
Fahrenheit (°F): T(°F) = (9/5)T(°C) + 32.

A change of 1°C equals a change of 1 K (but a change of 1.8°F).

Thermal Equilibrium and Heat

Two bodies are in thermal equilibrium when there is no net flow of heat between them, equivalently, they have the same temperature. The Zeroth Law of Thermodynamics: if A is in thermal equilibrium with C and B is in thermal equilibrium with C, then A is in thermal equilibrium with B. This justifies the very concept of temperature.

Heat (Q) is energy in transit between systems due to a temperature difference. SI unit: joule (J). 1 calorie = 4.18 J. Heat is a process quantity (path-dependent), not a state function.

Internal energy

Internal energy U is the total kinetic + potential energy of all the molecules in a system. It is a state function, depends only on the present state, not the history. For an ideal gas U depends only on temperature.

Specific Heat, Calorimetry & Latent Heat

To raise the temperature of a substance (no phase change), the heat required is:

q = mcΔT

where m is mass, c is the specific heat (J/(kg·K) or J/(g·°C)), and ΔT is the temperature change. Water has an unusually large specific heat, c ≈ 4.18 J/(g·°C), important for biological temperature buffering. Calorimetry uses conservation of energy: heat lost by the hot object = heat gained by the cold object (q_lost + q_gained = 0).

During a phase change temperature stays constant while heat is added or removed; the heat goes into breaking/forming intermolecular bonds:

q = mL

where L is the latent heat: of fusion (L_f, solid↔liquid) or vaporization (L_v, liquid↔gas). For water L_f ≈ 334 J/g and L_v ≈ 2260 J/g. On a heating curve, sloped segments use q = mcΔT and flat (plateau) segments use q = mL.

Biological context. Sweating cools the body through the large latent heat of vaporization of water, evaporating a small mass of sweat removes a large amount of heat.

Work in Thermodynamics

When a gas changes volume, it does work. On a PV diagram the work done by the gas equals the area under the process curve (no calculus needed for MCAT). For the common constant-pressure case:

W = PΔV = P(V₂ − V₁)

Special cases:

Isobaric (constant P): W = P·ΔV = P(V₂ − V₁).
Isochoric (constant V): W = 0 (no volume change).
Isothermal (constant T): W = nRT·ln(V₂/V₁).
Adiabatic (Q = 0): PV^γ = constant; W = (P₁V₁ − P₂V₂)/(γ − 1).

Sign convention: W > 0 when work is done by the gas (expansion); W < 0 when work is done on the gas (compression).

First Law of Thermodynamics

Statement: heat supplied to a system is used to increase the internal energy of the system and/or to do work by the system on its surroundings.

ΔU = Q − W

This is simply conservation of energy. Sign conventions:

Q > 0: heat added to the system; Q < 0: heat lost.
W > 0: work done by the system (expansion); W < 0: work done on the system.

First law in special processes

Thermodynamic processes, what's constant, what's zero
Process	Constant	Q	W (by gas)	ΔU	P-V relation
Isothermal	T	Q = W	nRT·ln(V₂/V₁)	0	PV = nRT (Boyle's law)
Isobaric	P	nC_pΔT	P·ΔV	nC_vΔT	V/T = const (Charles's)
Isochoric	V	nC_vΔT = ΔU	0	nC_vΔT	P/T = const (Gay-Lussac)
Adiabatic	Q = 0	0	(P₁V₁ − P₂V₂) / (γ − 1)	−W	PV^γ = const
Cyclic	Returns to start	Q = W (= area of loop on PV)	Net = area enclosed	0	Closed curve on PV diagram

Common trap. Different textbooks write the First Law as ΔU = Q − W (chemistry/physics convention W = work BY gas) or ΔU = Q + W (some chemistry books, W = work ON gas). Always check which convention is in use; the MCAT physics convention uses ΔU = Q − W.

Molar Specific Heat of Gas

For a gas, the heat needed to raise the temperature of 1 mole by 1 K depends on the path of heating. Two principal molar specific heats are defined:

C_v: molar specific heat at constant volume: Q_v = nC_vΔT. Since W = 0 in an isochoric process, all heat goes to raising U: ΔU = nC_vΔT.
C_p: molar specific heat at constant pressure: Q_p = nC_pΔT. The gas also does PΔV work, so more heat is required for the same ΔT: C_p > C_v always.

Values for ideal gases

Monatomic gas (He, Ar): C_v = (3/2)R, C_p = (5/2)R, γ = 5/3 ≈ 1.67.
Diatomic gas (O₂, N₂): C_v = (5/2)R, C_p = (7/2)R, γ = 7/5 = 1.40.
Polyatomic gas: C_v ≈ 3R, C_p ≈ 4R, γ ≈ 4/3.

Ratio γ = C_p/C_v appears in the adiabatic equation PV^γ = constant.

Relation Cp − Cv = R (Mayer's Relation)

For 1 mole of an ideal gas:

C_p − C_v = R

Where R = 8.314 J/(mol·K) is the universal gas constant. Derivation idea: at constant V, all the heat raises U; at constant P, the same ΔT also requires PΔV = RΔT of work to be done on the surroundings, so more heat is required.

Memory aid. "Pumping a tyre needs more heat than a sealed bottle." Constant-pressure heating requires extra energy to perform expansion work. That extra is exactly R per mole per kelvin.

Second Law & Entropy

The Second Law of Thermodynamics states that the total entropy of an isolated system never decreases; spontaneous processes increase the total entropy of the universe (ΔS_universe > 0). Heat flows spontaneously from hot to cold, never the reverse without work input.

Entropy (S) is a state function measuring the dispersal/disorder of energy (J/K). For a reversible transfer of heat at temperature T: ΔS = q_rev/T. Entropy increases with melting, vaporization, dissolving, and rising temperature.

Common trap. A living organism can decrease its local entropy (building ordered molecules) only by exporting even more entropy to its surroundings, the entropy of the universe still rises. The Second Law is never violated.

Heat engines

A heat engine takes in heat Q_h from a hot reservoir at T_h, converts part of it into work W, and dumps Q_c to a cold reservoir at T_c. Efficiency:

η = W/Q_h = 1 − Q_c/Q_h. The maximum (Carnot) efficiency is η_max = 1 − T_c/T_h, with temperatures in kelvin. No real engine can exceed Carnot efficiency, a consequence of the Second Law.

Heat Transfer Modes

Three modes of heat transfer
Mode	Mechanism	Medium	Example
Conduction	Direct molecular collisions transfer kinetic energy	Solids (esp. metals)	Heat through a metal spoon; body core to skin
Convection	Bulk movement of a heated fluid carries energy	Liquids and gases	Boiling water; blood circulation distributing body heat
Radiation	Emission of electromagnetic waves; no medium needed	Any (incl. vacuum)	Sun's heat; body heat lost as infrared

The body regulates temperature using all three, plus evaporation of sweat (latent heat). Radiated power rises steeply with absolute temperature (Stefan–Boltzmann, P ∝ T⁴).

Worked MCQs

Five MCQs that capture the high-yield testing patterns for this chapter.

Q1. A gas absorbs 500 J of heat and does 200 J of work. The change in its internal energy is:

700 J
300 J
500 J
−300 J

First Law: ΔU = Q − W = 500 − 200 = 300 J.

Q2. For an ideal monatomic gas, the molar specific heat at constant volume is:

R
(3/2)R
(5/2)R
(7/2)R

A monatomic gas has only translational degrees of freedom (3). Equipartition gives U = (3/2)nRT, so C_v = (3/2)R.

Q3. In an isothermal process applied to an ideal gas:

ΔU = Q
Q = W
Q = 0
W = 0

For an ideal gas U depends only on T, so isothermal means ΔU = 0 and the First Law gives Q = W.

Q4. Heat is added at a steady rate to a beaker of ice initially at 0°C. While the ice is melting, the temperature of the mixture:

Rises steadily
Stays constant at 0°C
Falls, then rises
Rises only after half the ice melts

During a phase change temperature is constant; the heat (q = mL_f) goes into breaking intermolecular bonds, not raising kinetic energy. Temperature rises only once all the ice has melted.

Q5. In an adiabatic process:

Q = 0 and ΔU = 0
Q = 0 and ΔU = −W
W = 0 and Q = ΔU
Q = −W and ΔU = 0

Adiabatic means no heat exchange (Q = 0). First Law ⇒ ΔU = −W. Adiabatic expansion cools the gas; adiabatic compression heats it.

Quick Recap

T(K) = T(°C) + 273.15; use kelvin in all gas-law/efficiency formulas.
Calorimetry: q = mcΔT (temperature change); q = mL (phase change, T constant).
First Law: ΔU = Q − W (W = work done BY gas); PV work = area under PV curve.
Isothermal: ΔU = 0; Isochoric: W = 0; Adiabatic: Q = 0; Cyclic: ΔU = 0.
C_v = (3/2)R (monatomic), (5/2)R (diatomic); C_p − C_v = R (Mayer's).
Second Law: ΔS_universe > 0 for spontaneous processes; heat flows hot → cold.
Carnot (max) efficiency η = 1 − T_c/T_h.
Heat transfer: conduction, convection, radiation.

Test yourself. Take a timed Thermodynamics quiz or browse all Physics MCQs to lock these concepts in.