Thermal energy transfers

The **total random kinetic energy** of all the particles **plus** the **total intermolecular potential energy** of all the particles.

**Random KE** (the particles' motion) and **intermolecular PE** (energy in the forces between particles).

Both the **random KE** of the particles **and** the **intermolecular PE** (a real gas has weak forces, so the PE part is not zero).

The **average random kinetic energy** of the particles (not the potential energy).

During a **change of state** (melting, boiling) — the spacing of the particles changes there.

The particles are packed **closer together** in the solid, so there is **more mass per volume**.

$\rho = \dfrac{m}{V}$ — mass ÷ volume. **Given** in the data booklet.

**kg m⁻³** (kilograms per cubic metre).

About **4 °C** — water's density anomaly.

Ice (and water below 4 °C) is **less dense** than water at 4 °C, so it rises and floats.

Ponds freeze **top-down**; the ice insulates the ≈4 °C water below, so fish survive the winter.

A **real gas** has KE **and** intermolecular PE; an **ideal gas** is modelled with no forces, so its internal energy is the **KE only**.

**298 K** (T/K = θ/°C + 273). A change of 25 °C is also a change of 25 K.

The **average random kinetic energy** of the particles: **Ē_k = 3/2 k_B T**.

The **temperature difference**: from higher to lower temperature, until thermal equilibrium (not from more internal energy to less).

Spacing ≈ **∛V** (each atom sits in a small cube).

The **energy needed to raise the temperature of 1 kg of a substance by 1 degree** (1 K). Unit: J kg⁻¹ K⁻¹.

**J kg⁻¹ K⁻¹** (joules per kilogram per kelvin: the energy to raise 1 kg by 1 K, i.e. 1 °C).

$Q = mc\Delta T$ — mass × specific heat capacity × temperature change. **Given** in the data booklet.

The temperature **change** = final temperature − start temperature (Δ means 'change in').

**No** — a change of 1 K is the same size as a change of 1 degree C, so either works. Never convert ΔT to kelvin.

$c = \dfrac{Q}{m\,\Delta T}$ — energy ÷ (mass × temperature change).

$m = \dfrac{Q}{c\,\Delta T}$.

Is **hard to heat** — it needs lots of energy per degree, so it warms and cools **slowly** (like water).

It has a **very large** specific heat capacity (about 4200 J kg⁻¹ K⁻¹), so it absorbs a lot of energy with only a small temperature rise.

During a **change of state** (melting/boiling), where the temperature stays constant — use $Q = mL$ instead.

Putting the **actual temperature** into ΔT instead of the **change** (final − start).

**Q = Pt = IVt**, with t in seconds.

Energy lost by the hot body = energy gained by the cold body: **m₁c₁ΔT₁ = m₂c₂ΔT₂**.

Energy is lost to the surroundings, so the block gains less than IVt, but all of IVt is used as Q.

The added energy goes into **breaking the bonds** between particles (latent heat), not into their kinetic energy — so the temperature does not change.

The **energy needed to change the state of 1 kg** of a substance with **no temperature change**. Unit: J kg⁻¹.

$Q = mL$ — energy = mass × specific latent heat. **Given** in the data booklet. Used for the flat parts (state change).

$Q = mc\Delta T$ — mass × specific heat capacity × temperature change. **Given** in the data booklet. Used for the sloping parts.

**Fusion (Lf)** = melting/freezing. **Vaporisation (Lv)** = boiling/condensing. For one substance, **Lv ≫ Lf**.

A **state change** (melting or boiling) at **constant temperature** — use $Q = mL$.

The **temperature is changing** (warming or cooling) — use $Q = mc\Delta T$.

Vaporising fully separates the particles, needing far more energy than melting (Lv ≫ Lf), so it takes longer at a steady heating rate.

**Energy lost by the hot object = energy gained by the cold object.** Add one Q-term per step (warm, melt, warm…).

Use a **separate Q-term for each step**: $Q = mc\Delta T$ for each temperature change and $Q = mL$ for each state change, then add them.

Some thermal energy is **lost to the surroundings** or absorbed by the **container**, which the ideal 'no losses' calculation ignores.

$Q = mL = 0.50 \times 3.3\times10^{5} = 1.65\times10^{5}$ J (≈ 1.7 × 10⁵ J).

Average **kinetic energy constant** (constant temperature); **potential energy** and internal energy **rise**.

Evaporation: fastest particles escape from the **surface** at any temperature, cooling the liquid. Boiling: throughout the liquid, at the boiling point.

The same power for both parts: **mcΔT/t₁ = m′L/t₂**, so P cancels.

**Conduction**, **convection** and **radiation**. Heat always flows from hotter to colder.

Faster-vibrating hot particles jostle their cooler neighbours, passing **energy** along while the particles stay put. In metals, free **electrons** also carry it (so metals conduct best).

The **hot fluid itself** moves: warmed fluid expands, becomes less dense and **rises**, carrying its energy with it (only in liquids and gases).

As **infrared electromagnetic waves**, needing **no material** — so it is the only method that works through a **vacuum** (e.g. the Sun → Earth).

**Radiation** only — conduction and convection both need particles/material.

$\dfrac{\Delta Q}{\Delta t} = kA\dfrac{\Delta T}{\Delta x}$ — rate = conductivity × area × (temperature difference ÷ thickness). **Given** in the data booklet.

The **watt** (W), i.e. joules per second (J s⁻¹) — it is a rate of energy transfer.

A bigger thickness **Δx** (on the bottom) **slows** conduction: rate ∝ 1 ÷ Δx, so doubling the thickness halves the rate.

A larger conductivity **k**, larger area **A**, or a larger temperature difference **ΔT**.

The object cools toward room temperature, so the **temperature difference** driving the heat loss shrinks — a smaller difference means a slower rate, i.e. a flatter graph.

They contain **free electrons** that move quickly through the metal and carry thermal energy, on top of the usual particle-to-particle vibration.

From a **hotter** region to a **colder** one, until they reach the same temperature (thermal equilibrium).

Heated fluid **expands, becomes less dense and rises**; cooler, denser fluid sinks to take its place.

The **rate of energy flow**: k₁(T₁ − T) = k₂(T − T₂) for bars of equal size.

Rate ∝ A ∝ **d²**: double the diameter, four times the rate.