Thermal properties of matter

IB Physics SL· 15 min read

1. Thermal Capacity and Specific Heat Capacity★★☆☆☆⏱ 5 min

📘 Definition

Thermal capacity

C

An extensive property (depends on amount of substance) that describes the total thermal energy needed to raise a given object's temperature by 1 K.

Example:

Relationship to specific heat capacity: $C = m c$ , where $m$ is mass and $c$ is specific heat capacity.

📘 Definition

Specific heat capacity

c

An intensive property (characteristic of the material, independent of amount) that describes the thermal energy needed to raise 1 kg of the substance by 1 K.

Example:

Pure water has $c = 4180 J kg^{- 1} K^{- 1}$

The thermal energy transferred to change the temperature of a mass $m$ by $Δ T$ is given by the formula $Q = m c Δ T$ . Note that a temperature change in degrees Celsius is equal to the same change in Kelvin, so unit conversion for $Δ T$ is not required.

📐 Worked Example

Calculate the thermal energy required to heat 0.5 kg of aluminium from 20 °C to 100 °C, given $c_{aluminium} = 900 J kg^{- 1} K^{- 1}$ .

1
1. Calculate the temperature change $Δ T$ :
2
$\Delta T = 100 - 20 = 80 \text{ K}$
3
1. Substitute into the heat transfer formula:
4
$Q = mc\Delta T = 0.5 \times 900 \times 80$
5
1. Calculate the final result:
6
$Q = 36000 \text{ J} = 36 \text{ kJ}$

2. Specific Latent Heat★★★☆☆⏱ 4 min

📘 Definition

Specific latent heat

L

The thermal energy required to change the phase of 1 kg of a substance at constant temperature. Latent heat of fusion ( $L_{f}$ ) describes melting/freezing, latent heat of vaporization ( $L_{v}$ ) describes boiling/condensation.

During a phase change, thermal energy changes the potential energy of particles (breaking or forming intermolecular bonds) rather than increasing average kinetic energy, so temperature remains constant. The total energy for a phase change of mass $m$ is $Q = m L$ .

📐 Worked Example

Calculate the energy required to melt 250 g of ice at 0 °C, given $L_{f, ice} = 3.34 \times 1 0^{5} J kg^{- 1}$ .

1
1. Convert mass to kilograms:
2
$m = 250 \text{ g} = 0.25 \text{ kg}$
3
1. Substitute into the latent heat formula:
4
$Q = mL_f = 0.25 \times 3.34 \times 10^5$
5
1. Calculate the result:
6
$Q = 83500 \text{ J} = 83.5 \text{ kJ}$

3. Mixed Equilibrium Problems★★★★☆⏱ 6 min

Many exam problems combine temperature changes and phase changes. By the principle of conservation of energy, assuming no heat is lost to the surroundings, the total thermal energy lost by a hot object equals the total thermal energy gained by a cold object.

📐 Worked Example

A 100 g copper block at 100 °C is placed into 200 g of water at 20 °C. Calculate the final equilibrium temperature, given $c_{copper} = 385 J kg^{- 1} K^{- 1}$ , $c_{water} = 4180 J kg^{- 1} K^{- 1}$ .

1
1. Let final temperature = $T$ , convert masses to kg: $m_{c} = 0.1 kg$ , $m_{w} = 0.2 kg$
2
1. Write energy lost by copper and energy gained by water:
3
$Q_{\text{lost}} = m_c c_c (100 - T), \quad Q_{\text{gained}} = m_w c_w (T - 20)$
4
1. Equate energies (conservation of energy):
5
$0.1 \times 385 \times (100 - T) = 0.2 \times 4180 \times (T - 20)$
6
1. Expand and rearrange to solve for $T$ :
7
$3850 - 38.5T = 836T - 16720 \\ 20570 = 874.5T \\ T \approx 23.5 ^\circ\text{C}$

4. Thermal Expansion (Particulate Explanation)★★☆☆☆⏱ 3 min

Most substances expand when heated because increasing average kinetic energy of particles increases the average spacing between them. Water is a key exception: it expands when it freezes, making ice less dense than liquid water (which is why ice floats).

✓ Quick check

Test your understanding of the particulate explanation:

Why does a solid expand when heated at constant pressure?
- A: The size of individual particles increases
- B: The average distance between particles increases
- C: The mass of particles increases
- D: The number of particles increases
Reveal answer
B —
Correct. Individual particles do not change size when heated; only the spacing between them increases as kinetic energy rises.

5. Common Pitfalls

Wrong move:

Forgetting to convert mass from grams to kilograms

Why:

Almost all specific heat and latent heat values are given per kilogram, so using grams gives a result 1000 times too small

Correct move:

Always convert mass to kilograms before substituting into $Q = m c Δ T$ or $Q = m L$

Wrong move:

Forgetting to add latent heat when heating through a melting/boiling point

Why:

Students often calculate only the total temperature change, ignoring the energy required for the phase transition

Correct move:

Add a separate $Q = m L$ term for any phase change that occurs between the initial and final temperature

Wrong move:

Mixing up thermal capacity and specific heat capacity

Why:

Thermal capacity is for the entire object, while specific heat capacity is per unit mass, leading to wrong values

Correct move:

Check units: thermal capacity has units $J K^{- 1}$ , specific heat has $J kg^{- 1} K^{- 1}$ , use $C = m c$ to convert between them

Wrong move:

Getting temperature differences reversed in equilibrium problems

Why:

This leads to a negative final temperature which is impossible

Correct move:

Always use (higher temperature - lower temperature) for both energy lost and energy gained, so both values are positive before equating

Wrong move:

Assuming temperature changes during a phase change

Why:

Students often incorrectly use $Q = m c Δ T$ for a phase change step

Correct move:

Temperature is constant during phase change, so always use $Q = m L$ for phase transitions

6. Quick Reference Cheatsheet

Quantity	Symbol	Formula	Units
Thermal Capacity	$C$	$C = m c$	$J K^{- 1}$
Specific Heat Capacity	$c$	$Q = m c Δ T$	$J kg^{- 1} K^{- 1}$
Specific Latent Heat	$L$	$Q = m L$	$J kg^{- 1}$
Energy Equilibrium		$Q_{lost} = Q_{gained}$
Temperature Change	$Δ T$	$Δ T (^{\circ} C) = Δ T (K)$	$K$ or $^{\circ} C$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 1
Specific heat multiple choice
2024 · 2
Latent heat calculation problem

What's Next

Understanding thermal properties of matter is a foundational skill for IB Physics SL, underpinning topics from thermodynamics to atmospheric energy transfer. This sub-topic connects everyday macroscopic observations (like ice floating or metal feeling cold) to core particulate theory, a unifying theme of the IB curriculum. Mastery of the calculations here is critical for both Paper 1 multiple choice and Paper 2 structured questions, as problems on thermal properties appear regularly in exams. The concepts you have learned here will now be extended to study heat transfer mechanisms, ideal gas behaviour, and kinetic theory.

→Wave behaviour
→Simple Harmonic Motion