Thermal Energy Transfer

Internal energy, the first law of thermodynamics, specific heat capacity and latent heat

SPEC 3.6.2.1 • THERMAL PHYSICS

Internal Energy

Definition

Internal energy is the sum of the randomly distributed kinetic energies and potential energies of all the molecules in a system.

Kinetic Energy Component

Due to random motion of molecules. Directly related to temperature.

Potential Energy Component

Due to intermolecular forces between molecules. Related to separation of molecules.

Key Concept

For an ideal gas, there are no intermolecular forces, so internal energy consists only of kinetic energy. For real gases and all other substances, both components contribute.

Internal Energy: Molecular Detail

At Equilibrium (r₀)

Force = 0 (point M) and PE is at minimum. PE is negative because work must be done to separate molecules to infinity (where PE = 0).

Kinetic Energy by State

Solid: Vibrational | Liquid: Vibrational + translational | Gas: Translational

Changing Internal Energy

The internal energy of a system can be increased by:

Heating (Q)

Transferring thermal energy to the system from a hotter object or surroundings.

Doing Work (W)

Work done on the system, e.g., compressing a gas or friction.

Exam Tip

Internal energy can increase even without a temperature rise — for example, during a change of state where energy increases the potential energy component (breaking intermolecular bonds).

First Law of Thermodynamics

First Law

The change in internal energy of a system equals the sum of the energy transferred by heating and the work done on the system.

Δ U = Q + W

Sign Convention (AQA)

QuantityPositive (+)Negative (−)

Q

Heat transferred TO systemHeat transferred FROM system

W

Work done ON systemWork done BY system

Δ U

Internal energy increasesInternal energy decreases

Common Mistake

Some textbooks use ΔU = Q − W where W is work done BY the system. AQA uses ΔU = Q + W where W is work done ON the system. Always check sign conventions!

Worked Example01

A gas in a cylinder is heated, receiving 850 J of thermal energy. The gas expands and does 320 J of work on the piston. Calculate the change in internal energy of the gas.

1

Identify values with correct signs:

Q = +850 J (heat transferred TO system)

W = −320 J (work done BY system, so negative)

2

Apply first law:

Δ U = Q + W

3

Substitute:

Δ U = 850 + (- 320) = 850 - 320

Answer: ΔU = +530 J (internal energy increases)

Isothermal Changes

Isothermal Process

An isothermal change occurs at constant temperature. Since temperature is constant, there is no change in internal energy: ΔU = 0

First Law for Isothermal Process

If ΔU = 0, then from ΔU = Q + W:

0 = Q + W \Rightarrow Q = - W

This means: heat supplied to the system = work done by the system

Physical Meaning

In an isothermal expansion, the gas does work on its surroundings. To keep temperature constant, an equal amount of heat must flow INTO the gas. The energy "passes through" the gas without changing its internal energy.

Specific Heat Capacity

Definition

Specific heat capacity (c) is the energy required to raise the temperature of 1 kg of a substance by 1 K (or 1°C), without changing its state.

E = m c Δ θ

Units

c has units of J kg⁻¹ K⁻¹ (or J kg⁻¹ °C⁻¹)

Where: E = energy transferred (J), m = mass (kg), Δθ = temperature change (K or °C)

MaterialSpecific Heat Capacity / J kg⁻¹ K⁻¹

Water4200

Copper385

Aluminium900

Ice2100

Note

A temperature change of 1 K is identical to a temperature change of 1°C, so either unit works in the formula.

Worked Example02

A kettle has a power of 2.0 kW and contains 0.80 kg of water at 20°C. How long will it take for the water to reach 100°C? (c_water = 4200 J kg⁻¹ K⁻¹)

1

Calculate temperature change: Δθ = 100 − 20 = 80 K

2

Calculate energy required:

E = m c Δ θ

E = 0.80 \times 4200 \times 80 = 268800

J

3

Use power equation:

P = \frac{E}{t}

→

t = \frac{E}{P}

t = \frac{268 800}{2000} = 134.4

s

Answer: t = 134 s (or 2 min 14 s)

Specific Latent Heat

Definition

Specific latent heat (L) is the energy required to change the state of 1 kg of a substance without changing its temperature.

E = m L

Latent Heat of Fusion (L_f)

Energy to change between solid and liquid at the melting point.

Water: L_f = 3.34 × 10⁵ J kg⁻¹

Latent Heat of Vaporisation (L_v)

Energy to change between liquid and gas at the boiling point.

Water: L_v = 2.26 × 10⁶ J kg⁻¹

Why No Temperature Change?

During a change of state, energy supplied increases the potential energy of molecules (breaking/forming intermolecular bonds) rather than their kinetic energy, so temperature stays constant.

Heating Curves

Reading Heating Curves

Sloped sections: Temperature changes — use E = mcΔθ (different c values for each state)

Flat sections: State changes — use E = mL (temperature constant, PE increasing)

Worked Example03

Calculate the total energy required to convert 0.40 kg of ice at −10°C to steam at 100°C.

Data: c_ice = 2100 J kg⁻¹ K⁻¹, c_water = 4200 J kg⁻¹ K⁻¹, L_f = 3.34 × 10⁵ J kg⁻¹, L_v = 2.26 × 10⁶ J kg⁻¹

1

Heat ice −10°C → 0°C:

E_{1} = 0.40 \times 2100 \times 10 = 8400

J

2

Melt ice at 0°C:

E_{2} = 0.40 \times 3.34 \times 1 0^{5} = 133600

J

3

Heat water 0°C → 100°C:

E_{3} = 0.40 \times 4200 \times 100 = 168000

J

4

Vaporise water at 100°C:

E_{4} = 0.40 \times 2.26 \times 1 0^{6} = 904000

J

5

Total:

E = 8400 + 133600 + 168000 + 904000 = 1214000

J

Answer: E = 1.21 MJ

Thermal Equilibrium

Principle

When two objects at different temperatures are placed in thermal contact in an insulated system, energy flows from the hotter to the cooler object until they reach the same final temperature.Energy lost by hot object = Energy gained by cold object

Method of Mixtures

For objects reaching thermal equilibrium:

m_{1} c_{1} (T_{1} - T_{f}) = m_{2} c_{2} (T_{f} - T_{2})

Where T_f is the final equilibrium temperature.

If a phase change occurs (e.g., ice melting), include the latent heat term.

Common Mistake

Forgetting to convert grams to kilograms, or assuming all ice melts when there may not be enough energy available.

Worked Example04

An ice cube of mass 0.025 kg at 0°C is dropped into 0.15 kg of water at 25°C in an insulated container. Calculate the final temperature.

Data: c_water = 4200 J kg⁻¹ K⁻¹, L_f = 3.34 × 10⁵ J kg⁻¹

1

Energy to melt ice:

E_{m e lt} = 0.025 \times 3.34 \times 1 0^{5} = 8350

J

2

Energy for melted ice to warm 0°C → T:

E_{i ce} = 0.025 \times 4200 \times T = 105 T

J

3

Energy lost by warm water 25°C → T:

E_{w a t er} = 0.15 \times 4200 \times (25 - T) = 630 (25 - T)

J

4

Energy balance:

8350 + 105 T = 630 (25 - T)

8350 + 105 T = 15750 - 630 T

735 T = 7400

Answer: T = 10.1°C

Worked Example05

A copper block of mass 0.50 kg at 95°C is placed in 0.20 kg of water at 15°C in an insulated container. Calculate the final temperature.

Data: c_copper = 385 J kg⁻¹ K⁻¹, c_water = 4200 J kg⁻¹ K⁻¹

1

Heat lost by copper (95°C → T):

E_{C u} = 0.50 \times 385 \times (95 - T) = 192.5 (95 - T)

J

2

Heat gained by water (15°C → T):

E_{w a t er} = 0.20 \times 4200 \times (T - 15) = 840 (T - 15)

J

3

Energy balance:

192.5 (95 - T) = 840 (T - 15)

18287.5 - 192.5 T = 840 T - 12600

30887.5 = 1032.5 T

Answer: T = 29.9°C ≈ 30°C

Worked Example06

Steam at 100°C is passed into 0.50 kg of water at 20°C until the water reaches 80°C. Calculate the mass of steam condensed.

Data: c_water = 4200 J kg⁻¹ K⁻¹, L_v = 2.26 × 10⁶ J kg⁻¹

1

Heat gained by cold water (20°C → 80°C):

E_{w a t er} = 0.50 \times 4200 \times 60 = 126000

J

2

Let mass of steam = m. Steam condenses at 100°C:

E_{co n d e n se} = m \times 2.26 \times 1 0^{6}

J

3

Condensed water cools (100°C → 80°C):

E_{coo l} = m \times 4200 \times 20 = 84000 m

J

4

Energy balance:

2.26 \times 1 0^{6} m + 84000 m = 126000

2.344 \times 1 0^{6} m = 126000

Answer: m = 0.054 kg = 54 g

Practice 1

A gas receives 1200 J of heat energy and has 450 J of work done on it during compression. Calculate the change in internal energy of the gas.

1

Q = +1200 J (heat in), W = +450 J (work on system)

2

Δ U = Q + W = 1200 + 450

Answer: ΔU = +1650 J

Practice 2

A 150 W heater is used to melt ice at 0°C. Calculate the time taken to melt 0.30 kg of ice completely. (L_f = 3.34 × 10⁵ J kg⁻¹)

1

Energy to melt ice:

E = m L_{f} = 0.30 \times 3.34 \times 1 0^{5} = 100200

J

2

Time:

t = \frac{E}{P} = \frac{100200}{150} = 668

s

Answer: t = 668 s (or 11.1 minutes)

Practice 3

A lead bullet of mass 0.015 kg travelling at 200 m s⁻¹ is stopped by a target. Assuming 80% of the kinetic energy is converted to internal energy of the bullet, calculate the temperature rise. (c_lead = 130 J kg⁻¹ K⁻¹)

1

Kinetic energy:

K E = \frac{1}{2} m v^{2} = \frac{1}{2} \times 0.015 \times 20 0^{2} = 300

J

2

Energy to bullet:

E = 0.80 \times 300 = 240

J

3

Temperature rise:

Δ θ = \frac{E}{m c} = \frac{240}{0.015 \times 130}

Answer: Δθ = 123 K (or 123°C)

Summary

Key Formulas

ConceptFormulaNotes

First Law of Thermodynamics

Δ U = Q + W

W = work ON system

Isothermal Process

Δ U = 0 \to Q = - W

Constant temperature

Specific Heat Capacity

E = m c Δ θ

c in J kg⁻¹ K⁻¹

Specific Latent Heat

E = m L

L in J kg⁻¹

Key Exam Skills

Apply correct sign convention for first law (Q and W positive when energy enters system)

Interpret heating curves — identify phases and apply correct formula

Combine SHC and SLH calculations for multi-stage heating/cooling problems

Set up and solve thermal equilibrium (method of mixtures) problems

Explain why temperature remains constant during state changes (PE increasing, not KE)