When 4.219 kJ of energy are added as heat to 36.0 grams of water at one bar, the temperature of the water increases from 10.00°C to 38.05°C. Calculate the heat capacity of the 36.0 grams of H_{2}O(l).
We use Equation 14.27. The value of ΔT is equal to 38.05°C – 10.00°C = 28.05°C. Because a kelvin and a Celsius degree are the same size (see sidebox), we have
c_{P}=\frac{q_{P}}{\Delta T} =\frac{4219\ J}{28.05\ K} = 150.4\ J·K^{-1}for the heat capacity of 36.0 grams of H_{2}O(l).