For a material heated without any change of state, energy transferred and temperature change are proportional, so the graph is a straight line. The gradient carries m × c, and that is how IGCSE Physics questions hide the specific heat capacity in a graph.
This lesson builds directly on using specific heat capacity with compatible units.
How does a slope connect to Q = m × c × Δθ?
Rearrange the equation as Q = (m × c) × Δθ. This has the shape y = gradient × x, with energy on the vertical axis and temperature change on the horizontal axis. So the gradient is m × c.
If the axes are the other way round, with temperature on the vertical axis, the gradient is Δθ ÷ Q, which is 1 ÷ (m × c). Always decide which quantity is on which axis before you calculate.
- Read the axis labels and units, including any kJ or ×10³ multipliers.
- Pick two points far apart on the straight part of the line.
- Take a change on each axis: rise = difference in y, run = difference in x.
- Divide rise by run, and keep the units of the quotient.
- Convert to c by dividing m × c by the mass in kg, or by taking the reciprocal first if temperature is on the vertical axis.
Worked example
A 0.50 kg sample of an unknown liquid is heated and the energy supplied is plotted against temperature. (Invented example data.) The line passes through (20 °C, 0 J) and (45 °C, 50 000 J). Find the specific heat capacity.
Step 1, axes: energy (J) is vertical and temperature (°C) is horizontal, so the gradient is m × c.
Step 2, changes: rise = 50 000 − 0 = 50 000 J. Run = 45 − 20 = 25 °C.
Step 3, gradient: 50 000 ÷ 25 = 2000 J/°C. So m × c = 2000 J/°C.
Step 4, c: c = 2000 ÷ 0.50 = 4000 J/kg °C.
Step 5, check: Q = 0.50 × 4000 × 25 = 50 000 J, which matches the last point on the graph.
The mistake to watch for
A frequent slip is to use the gradient of the wrong axis arrangement.
Mistaken answer: The same data is plotted with temperature vertical. The student finds a gradient of 25 ÷ 50 000 = 0.0005 and says c = 0.0005 ÷ 0.50 = 0.001 J/kg °C.
The gradient 0.0005 °C/J is the reciprocal of m × c, not m × c itself.
The correction is to invert first: 1 ÷ 0.0005 = 2000 J/°C, then divide by 0.50 kg to get 4000 J/kg °C. A sense check also exposes the error, because no ordinary material has a specific heat capacity as tiny as 0.001 J/kg °C.
Check yourself
1. A line of energy against temperature for a 0.30 kg block passes through (10 °C, 0 J) and (30 °C, 6000 J). Find c.
Show answer
Gradient = 6000 ÷ 20 = 300 J/°C = m × c. c = 300 ÷ 0.30 = 1000 J/kg °C.
2. Two liquids of equal mass receive the same energy. Liquid A rises by 30 °C and liquid B rises by 20 °C. Which has the larger specific heat capacity, and why?
Show answer
Liquid B. With the same Q and m, c = Q ÷ (m × Δθ). A smaller temperature rise means a larger c.
3. A graph of temperature (vertical) against energy (horizontal) for a 0.25 kg object has a gradient of 0.004 °C/J. Find c.
Show answer
m × c = 1 ÷ 0.004 = 250 J/°C. c = 250 ÷ 0.25 = 1000 J/kg °C.
Where this leads next
The next idea is what happens when the line stops being straight: distinguish a temperature change from a phase change. The heat calculations practice set mixes graph and equation questions.
If you can calculate but freeze when the same data is shown as a graph, that gap is a good use of online one-to-one Physics tuition.