Calculating heat energy

When a fuel burns or a reaction happens in water, the heat released (or absorbed) flows into (or out of) the surrounding water. We measure how much the water's temperature changes and use that to calculate the heat energy, Q.

The formula is:

$Q = m \times c \times \Delta T$

Each letter has an exact meaning — and a fixed unit:

Symbol	Meaning	Unit
Q	heat energy change	joules (J)
m	mass of WATER being heated	grams (g)
c	specific heat capacity of water	4.18 J/g/°C
ΔT	temperature change (final − initial)	°C

The single most important point: in this formula, m is the mass of the WATER, not the mass of the fuel that burns and not the mass of any solid that dissolves. The water is what gets heated, so the water's mass goes into the formula.

What is specific heat capacity? It is the energy needed to raise the temperature of 1 gram of a substance by 1 °C. For water, c = 4.18 J/g/°C — so every gram of water needs 4.18 J to warm up by 1 °C.

Note on units of c. You may see c written as 4.18 J/g/°C or 4.18 J/g/K. The numbers are identical because a change of 1 °C is the same size as a change of 1 K — so use whichever appears on your data.

Questions almost always give you the volume of water in cm³, but the formula needs the mass in grams. Luckily the conversion is easy:

$\textbf{1 cm}^3 \textbf{ of water} \approx \textbf{1 g}$

This is because water has a density of about 1 g/cm³. So:

• 100 cm³ of water → 100 g
• 250 cm³ of water → 250 g
• 50 cm³ of water → 50 g

Worked-in example. A question says "a spirit burner of ethanol is used to heat 200 cm³ of water". For the formula, m = 200 g.

The classic trap — choosing the wrong mass. A combustion question gives you TWO masses:

1. the mass of fuel burned (e.g. 0.92 g of ethanol), and
2. the mass (or volume) of water heated (e.g. 250 g).

For Q = mcΔT you must use the mass of WATER (250 g). The mass of fuel is used later (to find moles, in a different calculation) — never inside Q = mcΔT. Putting the fuel mass into Q = mcΔT is one of the most common ways students throw away marks.

For solutions (neutralisation / dissolving): when two solutions are mixed, add the volumes and treat the total as water:

• 50 cm³ acid + 50 cm³ alkali → m = 100 g.

We assume the solution has the same density and specific heat capacity as water, which is a fair approximation for dilute solutions.

Finding ΔT. The temperature change is always:

$\Delta T = T_{\text{final}} - T_{\text{initial}}$

For example, if water warms from 18.5 °C to 41.0 °C:

$\Delta T = 41.0 - 18.5 = 22.5\ °\text{C}$

• If the temperature rises, the reaction is exothermic (heat released into the water).
• If the temperature falls, the reaction is endothermic (heat absorbed from the water) — then take ΔT as the size of the drop.

Converting joules to kilojoules. The formula gives Q in joules (J), but answers are often wanted in kilojoules (kJ). There are 1000 J in 1 kJ, so:

$Q\ (\text{kJ}) = \dfrac{Q\ (\text{J})}{1000}$

• 8360 J = 8.36 kJ
• 20 900 J = 20.9 kJ
• 2842 J = 2.84 kJ

Always read what unit the question wants. If it asks for the answer in kJ, you lose a mark if you leave it in J (or forget to divide by 1000).

Sometimes the exam gives you Q and asks you to find the temperature change or the mass of water. You rearrange the formula:

To find ΔT (temperature change):

$\Delta T = \dfrac{Q}{m \times c}$

To find m (mass of water):

$m = \dfrac{Q}{c \times \Delta T}$

A quick way to remember the triangle: Q sits on top; m, c and ΔT sit underneath, multiplied together.

You want to find…	Use…
Q	Q = m × c × ΔT
ΔT	ΔT = Q ÷ (m × c)
m	m = Q ÷ (c × ΔT)

Tip: keep c = 4.18 in every rearrangement — only the position of c changes, never its value. Always show your rearranged formula first, then substitute the numbers — examiners award method marks even if your final arithmetic slips.