# How can I calculate specific heat capacity of water?

Dec 29, 2014

Specific heat represents the amount of heat required to increase the temperature of one gram of a substance by one degree Celsius.

$q = m \cdot c \cdot \Delta T$

So, if you know how much heat was added to a certain mass of water to increase its temperature by a number of degrees, you could calculate water's specific heat quite easily.

Let's assume 94.1 kJ were provided to 0.50 L of water to increase its temperature from 20.2 to 65.2 degrees Celsius. Since we know that water has a density of $1.0 \text{kg/L}$, we can determine its mass by

${m}_{w a t e r} = \rho \cdot V = 1.0 \frac{k g}{L} \cdot 0.50 L = 0.50 k g = 500.0 g$

So,

$c = \frac{q}{m \cdot \Delta T} = \frac{94.1 \cdot {10}^{3} J}{500.0 g \cdot {\left(65.2 - 20.2\right)}^{\circ} C} = 4.18 \frac{J}{g {\cdot}^{\circ} C}$

Usually, problems that ask you to calculate a substance's specific heat will provide such information (heat, $\Delta T$, mass).