Study Guide

Temperature change and heat capacity

IB Chemistry SL· 35 min read

1. Key Definitions and Core Relationship★★☆☆☆⏱ 10 min

📘 Definition

Heat Capacity

C=qΔTC = \frac{q}{\Delta T}

The total amount of heat energy required to change the temperature of a given sample of substance by 1 K (or 1°C). It is an extensive property that depends on the amount of substance.

Example:

A 50 g block of copper has a total heat capacity of ~19 J K⁻¹

📘 Definition

Specific Heat Capacity

c=qmΔTc = \frac{q}{m\Delta T}

The heat energy required to change the temperature of 1 gram of a substance by 1 K. It is an intensive property, characteristic of the substance itself.

Example:

Liquid water has a specific heat capacity of 4.18 J g⁻¹ K⁻¹

Combining these definitions gives the core equation for all heat transfer calculations relating heat to temperature change:

q=mcΔTq = mc\Delta T
📐 Worked Example

A 100 g sample of metal requires 450 J of heat to raise its temperature from 25°C to 45°C. Calculate the specific heat capacity of the metal.

  1. 1

    First calculate the temperature change ΔT (note that 1°C change = 1 K change):

  2. 2
    ΔT=TfinalTinitial=45C25C=20K\Delta T = T_{final} - T_{initial} = 45^\circ C - 25^\circ C = 20 K
  3. 3

    Rearrange the core equation to solve for :

  4. 4
    c=qmΔTc = \frac{q}{m \Delta T}
  5. 5

    Substitute the given values:

  6. 6
    c=450 J100 g×20 K=0.225 J g1K1c = \frac{450 \text{ J}}{100 \text{ g} \times 20 \text{ K}} = 0.225 \text{ J g}^{-1} \text{K}^{-1}

2. Applications to Simple Calorimetry★★★☆☆⏱ 15 min

In simple experimental calorimetry, heat released or absorbed by a chemical reaction is transferred to the surrounding solution (usually water) in an insulated container. For IB SL calculations, we assume no heat is lost to the surroundings, so:

qreaction=qsolution|q_{reaction}| = |q_{solution}|
📐 Worked Example

When 0.1 mol of NaOH dissolves in 200 g of water, the temperature rises from 22°C to 28.5°C. Calculate the heat released by dissolution (assume J g⁻¹ K⁻¹).

  1. 1

    Calculate the temperature change of the solution:

  2. 2
    ΔT=28.5C22.0C=6.5K\Delta T = 28.5^\circ C - 22.0^\circ C = 6.5 K
  3. 3

    Calculate heat gained by the solution using :

  4. 4
    qsolution=200 g×4.18 J g1K1×6.5K=5434J=5.43kJq_{solution} = 200 \text{ g} \times 4.18 \text{ J g}^{-1} K^{-1} \times 6.5 K = 5434 J = 5.43 kJ
  5. 5

    Heat gained by the solution equals heat released by the reaction. So heat released by dissolution is 5.43 kJ, and kJ.

✓ Quick check

Test your understanding of sign convention:

  1. The temperature of water in a calorimeter decreases during a reaction. What is the sign of for the reaction?

    • Negative

    • Positive

    • Zero

    Reveal answer
    1

    Correct! A temperature decrease means the reaction absorbs heat from the water, so it is endothermic and is positive.

3. Heat Capacity vs Specific Heat Capacity★★★☆☆⏱ 10 min

The most common confusion in this topic is mixing up total heat capacity and specific heat capacity. The table below summarises the key differences:

Property

Depends on mass?

Units

Core Equation

Heat capacity (C)

Yes

J K⁻¹

Specific heat capacity (c)

No

J g⁻¹ K⁻¹

📐 Worked Example

A 20 g sample of iron has a specific heat capacity of 0.45 J g⁻¹ K⁻¹. What is the total heat capacity of the 20 g sample, and what is the specific heat capacity of a 40 g sample of iron?

  1. 1

    Calculate total heat capacity for the 20 g sample:

  2. 2
    C=m×c=20 g×0.45 J g1K1=9 J K1C = m \times c = 20 \text{ g} \times 0.45 \text{ J g}^{-1} K^{-1} = 9 \text{ J K}^{-1}
  3. 3

    Specific heat capacity is an intensive property, so it does not change with mass. The 40 g sample therefore still has a specific heat capacity of 0.45 J g⁻¹ K⁻¹.

  4. 4

    The total heat capacity of the 40 g sample doubles, since it depends on mass:

  5. 5
    C=40 g×0.45 J g1K1=18 J K1C = 40 \text{ g} \times 0.45 \text{ J g}^{-1} K^{-1} = 18 \text{ J K}^{-1}

4. Common Pitfalls

Wrong move:

Mismatching mass units to specific heat capacity units

Why:

Most specific heat capacities for IB SL are given in J g⁻¹ K⁻¹, so mass in kg will give a result 1000x too small

Correct move:

Always check the units of before substituting, and convert mass to match the given units

Wrong move:

Getting the sign of wrong for exothermic reactions

Why:

Students often report a positive because temperature increased, but exothermic reactions release heat so their is negative

Correct move:

Remember: of solution increases → exothermic →

Wrong move:

Using the specific heat of the reactant instead of the solution

Why:

All heat transfer is assumed to go to the surrounding aqueous solution, not the reactant itself

Correct move:

Unless stated otherwise, use J g⁻¹ K⁻¹ for all aqueous solutions in IB SL problems

Wrong move:

Confusing heat capacity and specific heat capacity in calculations

Why:

Students forget to multiply/divide by mass when switching between the two properties

Correct move:

Always label your variables and check what quantity the question asks you to calculate

Wrong move:

Converting ΔT from Celsius to Kelvin incorrectly

Why:

Students add 273 to ΔT, but the change is the same for both scales

Correct move:

A ΔT of 1°C is exactly equal to a ΔT of 1 K, so no conversion of the difference is needed

5. Quick Reference Cheatsheet

Term

Symbol

Units

Relationship

Heat capacity

C

J K⁻¹

Specific heat capacity

c

J g⁻¹ K⁻¹

Heat transfer

q

J

Specific heat (water)

4.18 J g⁻¹ K⁻¹

Use for aqueous solutions

ΔT (C vs K)

K / °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.

  • 2025 · Paper 1

    Specific heat calculation

  • 2024 · Paper 2

    Calorimetry temperature change

What's Next

Mastery of temperature change and heat capacity is the foundation for all calorimetry and enthalpy calculations in IB Chemistry SL. The relationship is used directly to calculate enthalpy changes of reaction from experimental data, and is required for all subsequent topics in the energetics unit, including Hess's law and bond enthalpy calculations. This concept also underpins practical assessment questions on calorimetry experiments, which regularly appear in internal assessments and exam practicals.