Thermodynamics · Chapter 10

Heat and Temperature

Temperature measures the average kinetic energy of microscopic motion — heat is the transfer of that energy across a boundary.

Prerequisites

Energy & Work \cdot Basic mechanics \cdot Ideal Gas Law (helpful but not required

Learning Goals

Distinguish temperature from heat and convert between Celsius, Fahrenheit, and Kelvin scales.
$Apply Q = mc\DeltaT to calculate heat exchange and final equilibrium temperature in mixing prob$ lems.
Identify the three mechanisms of heat transfer and apply Fourier's and Stefan-Boltzmann's laws.
$Calculate latent heat for phase changes using Q = mL$ .
$Compute rms molecular speed from the kinetic theory formula v_{rms} = \sqrt(3k_BT/m$ .

10.1 Temperature and the Zeroth Law

Temperature is a measure of the average translational kinetic energy of the particles in a substance. We cannot observe individual molecular motion directly, but we can measure its macroscopic effect: the tendency to spontaneously exchange energy with surroundings.

Definition 10.1 — Zeroth Law of Thermodynamics

If two systems A and B are each in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other. This law — logically prior to the other three — is the basis for temperature as a meaningful, transitive quantity, and for the operation of thermometers.

The three common temperature scales relate by exact conversions:

T_K=T_C+273.15 \qquad T_F=\frac{9}{5}T_C+32

(10.1)

The Kelvin scale is the fundamental one — it starts at absolute zero, the temperature at which thermal motion would theoretically cease. All thermodynamic formulas require Kelvin.

10.2 Heat Transfer and Specific Heat

Heat Q is energy in transit — it flows from a hotter body to a cooler one until thermal equilibrium is reached. Heat is not stored; temperature is. When Q joules of heat enter a substance of mass m, the temperature change ΔT depends on the material:

Q=mc\Delta T

(10.2)

where c is the specific heat capacity (J/kg·K) — how much energy per kilogram per degree it takes to warm that material. Water has an unusually high c = 4186 J/kg·K, which is why oceans moderate coastal climates. Metals are far lower (aluminum: 900, iron: 450).

Example 10.1 — Mixing Hot and Cold Water

200 g of water at 80°C is mixed with 300 g of water at 20°C in an insulated container. Find the final temperature.

Heat balance:

Heat lost by hot = heat gained by cold: m_{1}c(T_{1}-T_{f}) = m_{2}c(T_{f}-T_{2}

Solve:

0.2(80-T_{f}) = 0.3(T_{f}-20) \to 16-0.2T_f = 0.3T_f-6 \to 22 = 0.5T_f \to T_{f} =

44°C

Check:

Weighted average: (0.2\times80 + 0.3\times20)/(0.2+0.3) = (16+6)/0.5 = 44°C ✓

10.3 Heat Transfer Mechanisms

Heat moves by three mechanisms:

Conduction — through direct molecular contact. Heat flux: Q/t = kA(ΔT/d), where k is thermal conductivity (W/m·K), A is area, and d is thickness.
Convection — by bulk fluid motion. Hot fluid rises, carrying energy. Responsible for ocean circulation and atmospheric weather.
Radiation — via electromagnetic waves (photons), requiring no medium. A blackbody emits power P = σT⁴A (Stefan-Boltzmann law, σ = 5.67×10⁻⁸ W/m²·K⁴).

P_\mathrm{conduction}=kA\frac{\Delta T}{d} \qquad P_\mathrm{radiation}=\varepsilon\sigma AT^4

(10.3)

10.4 Phase Changes and Latent Heat

When a substance changes phase (solid ↔ liquid ↔ gas), energy is absorbed or released at constant temperature. This energy goes into rearranging molecular bonds, not increasing kinetic energy:

Q=mL

(10.4)

where L is the latent heat (J/kg). For water: L_fusion = 334 kJ/kg (melting ice), L_vaporization = 2257 kJ/kg (boiling water). The enormous L_vap is why sweating cools you so effectively — evaporating 1 g of sweat removes 2257 J from your skin.

Definition 10.2 — Thermal Equilibrium Condition

When an isolated system reaches thermal equilibrium, all heat exchange has ceased. For two objects mixing: the total enthalpy is conserved (no work done, no phase change):

\sum m_{i}c_{i}(Tᶠ - T_{i}) = 0

Each term is positive if the object gains heat, negative if it loses heat. The sum is exactly zero for a perfectly insulated system.

10.5 Kinetic Theory of Temperature

At the microscopic level, temperature is a measure of average translational kinetic energy per particle. For an ideal gas of N molecules, the equipartition theorem gives:

\frac{1}{2}mv_\mathrm{rms}^2=\frac{3}{2}k_BT \qquad v_\mathrm{rms}=\sqrt{\frac{3k_BT}{m}}

(10.5)

where k_B = 1.38×10⁻²³ J/K is Boltzmann's constant. At room temperature (T = 293 K), nitrogen molecules move at v_rms ≈ 511 m/s — faster than a rifle bullet. The simulation below shows this directly: faster particles appear redder.

Figure 10.1.

Kinetic gas simulation. Each dot is a molecule. Color encodes speed (blue=slow, red=fast

. Raise the temperature to watch the speed distribution shift. Compress the volume (move the piston) to see pressure increase — Boyle's law in action.

Example 10.2 — RMS Speed of Oxygen

$Find the rms speed of O_{2} molecules at T = 300 K. (m_O_{2} = 32 u = 5.31\times10^{-26} kg$

Formula:

v_{rms} = \sqrt(3k_BT/m

Calculate:

v_{rms} = \sqrt(3 \times 1.38\times10^{-23} \times 300 / 5.31\times10^{-26}) = \sqrt(2.34\times10^{5}) =

484 m/s

Definition 10.3 — Common Traps

Heat is not temperature: heat is energy crossing a boundary; temperature is a state variable.
Use Kelvin for proportional laws: Celsius differences are fine, but absolute temperature formulas need kelvin.
Phase changes happen at constant temperature: during melting or boiling, added energy changes phase before raising T.
Radiation depends on absolute temperature: $Stefan-Boltzmann uses T in kelvin and scales as T^{4}$

Exercises — 10.1–10.5 Heat and Temperature

Convert 100°C to Kelvin.

Straightforward

How much heat is required to warm 250 g of water from 20°C to 100°C? (c = 4186 J/kg\cdotK

Straightforward

100 g of water at 80°C is mixed with 100 g of water at 20°C. Find the final equilibrium temperature.

°C

Intermediate

How much heat is released when 100 g of steam at 100°C condenses to liquid water? (L_{vap} =

2.26\times10^{6} J/kg

Intermediate

A 10 g lead bullet (c = 128 J/kg\cdotK) moving at 300 m/s embeds in a 0.5 kg wood block (c = 1

700 J/kg\cdotK). Assuming all kinetic energy converts to heat, estimate the temperature rise o

f the bullet.

Challenging

Key Takeaways

Temperature is average molecular KE; heat is energy in transfer — they are not the same thing.
$T(K) = T(°C) + 273.15 — always use Kelvin in thermodynamic formulas$ .
$Q = mc\DeltaT for sensible heat; Q = mL for phase changes at constant temperature$ .
Three mechanisms: conduction (contact), convection (fluid flow), radiation (EM waves).
$Kinetic theory: v_{rms} = \sqrt(3k_BT/m) — molecules at room temperature move at hundreds of m/s$ .