Worked example
Two points read off the straight part of a temperature against energy line: (0, 20.0) and (40 000 J, 64.0 °C).
- Change in y = 64 − 20 = 44 °C
- Change in x = 40000 − 0 = 40000 J
- Gradient = 44 ÷ 40000 = 0.00110 °C/J
The y-intercept is 20.0 °C, the starting temperature. In the specific heat capacity practical, c = 1 ÷ (mass × gradient).