# Temperature change and heat capacity

> IB Chemistry SL · Reactivity 2: Energetics and Kinetics
> Source: https://www.owlsprep.com/study/ib-chemistry-sl-u5-temperature-change-and-heat-capacity/

This sub-topic explains the relationship between heat energy transfer, temperature change, and heat capacity. You will learn to apply the core $q = mc\Delta T$ equation to solve common calorimetry problems for IB Chemistry SL.

**Prerequisites:** [Basic understanding of heat and temperature](https://www.owlsprep.com/study/ib-chemistry-sl-u5-introduction-to-energetics/); Unit conversion for energy and temperature

## Learning objectives

- Distinguish between heat capacity and specific heat capacity
- Calculate temperature change and heat transfer for thermal processes
- Apply the $q = mc\Delta T$ equation to simple calorimetry problems
- Interpret the sign of $q$ for exothermic and endothermic processes

## Key Definitions and Core Relationship

**Heat Capacity** — 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.

*Notation:* C = \frac{q}{\Delta T}

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

**Specific Heat Capacity** — 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.

*Notation:* c = \frac{q}{m\Delta T}

*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:

$$q = mc\Delta T$$

**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. First calculate the temperature change ΔT (note that 1°C change = 1 K change):
2. $$\Delta T = T_{final} - T_{initial} = 45^\circ C - 25^\circ C = 20 K$$
3. Rearrange the core equation to solve for $c$:
4. $$c = \frac{q}{m \Delta T}$$
5. Substitute the given values:
6. $$c = \frac{450 \text{ J}}{100 \text{ g} \times 20 \text{ K}} = 0.225 \text{ J g}^{-1} \text{K}^{-1}$$

## Applications to Simple Calorimetry

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:

$$|q_{reaction}| = |q_{solution}|$$

> **tip**
>
> Sign convention: If the reaction temperature increases, the reaction is exothermic, so $q_{reaction}$ is negative. If temperature decreases, the reaction is endothermic, so $q_{reaction}$ is positive.

**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 $c_{solution} = 4.18$ J g⁻¹ K⁻¹).

1. Calculate the temperature change of the solution:
2. $$\Delta T = 28.5^\circ C - 22.0^\circ C = 6.5 K$$
3. Calculate heat gained by the solution using $q = mc\Delta T$:
4. $$q_{solution} = 200 \text{ g} \times 4.18 \text{ J g}^{-1} K^{-1} \times 6.5 K = 5434 J = 5.43 kJ$$
5. Heat gained by the solution equals heat released by the reaction. So heat released by dissolution is 5.43 kJ, and $q_{reaction} = -5.43$ kJ.

**Check your understanding**

Test your understanding of sign convention:

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

   - Negative
   - Positive
   - Zero

   *Answer:* Positive

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

## Heat Capacity vs Specific Heat Capacity

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 | $C = q/\Delta T$ |
| Specific heat capacity (c) | No | J g⁻¹ K⁻¹ | $c = q/(m\Delta T)$ |

**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. Calculate total heat capacity for the 20 g sample:
2. $$C = m \times c = 20 \text{ g} \times 0.45 \text{ J g}^{-1} K^{-1} = 9 \text{ J K}^{-1}$$
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. The total heat capacity of the 40 g sample doubles, since it depends on mass:
5. $$C = 40 \text{ g} \times 0.45 \text{ J g}^{-1} K^{-1} = 18 \text{ J K}^{-1}$$

## Common pitfalls

- **Wrong:** Mismatching mass units to specific heat capacity units
  - Why it fails: 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: Always check the units of $c$ before substituting, and convert mass to match the given units
- **Wrong:** Getting the sign of $q_{reaction}$ wrong for exothermic reactions
  - Why it fails: Students often report a positive $q$ because temperature increased, but exothermic reactions release heat so their $q$ is negative
  - Correct: Remember: $\Delta T$ of solution increases → exothermic → $q_{reaction} < 0$
- **Wrong:** Using the specific heat of the reactant instead of the solution
  - Why it fails: All heat transfer is assumed to go to the surrounding aqueous solution, not the reactant itself
  - Correct: Unless stated otherwise, use $c = 4.18$ J g⁻¹ K⁻¹ for all aqueous solutions in IB SL problems
- **Wrong:** Confusing heat capacity and specific heat capacity in calculations
  - Why it fails: Students forget to multiply/divide by mass when switching between the two properties
  - Correct: Always label your variables and check what quantity the question asks you to calculate
- **Wrong:** Converting ΔT from Celsius to Kelvin incorrectly
  - Why it fails: Students add 273 to ΔT, but the change is the same for both scales
  - Correct: A ΔT of 1°C is exactly equal to a ΔT of 1 K, so no conversion of the difference is needed

## Cheatsheet

| Term | Symbol | Units | Relationship |
| --- | --- | --- | --- |
| Heat capacity | C | J K⁻¹ | $C = q/\Delta T$ |
| Specific heat capacity | c | J g⁻¹ K⁻¹ | $c = q/(m\Delta T)$ |
| Heat transfer | q | J | $q = mc\Delta T$ |
| Specific heat (water) | $c_{water}$ | 4.18 J g⁻¹ K⁻¹ | Use for aqueous solutions |
| ΔT (C vs K) | $\Delta T$ | K / °C | $\Delta T (K) = \Delta T (^\circ C)$ |

## What's next

Mastery of temperature change and heat capacity is the foundation for all calorimetry and enthalpy calculations in IB Chemistry SL. The $q = mc\Delta T$ 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.

- [Enthalpy of reaction](https://www.owlsprep.com/study/ib-chemistry-sl-u5-enthalpy-of-reaction/)
- [Hess's Law](https://www.owlsprep.com/study/ib-chemistry-sl-u5-hess-s-law/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ib-chemistry-sl-u5-temperature-change-and-heat-capacity/