Question

How can I calculate specific heat capacity of water?

Answer 1

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

`q = m * c * DeltaT`

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 "kg/L"`, we can determine its mass by

`m_(water) = rho * V = 1.0 (kg)/(L) * 0.50L = 0.50 kg = 500.0g`

So,

`c = q/(m * DeltaT) = (94.1 * 10^3 J)/(500.0g * (65.2-20.2)^@C) = 4.18 J/(g*^@C)`

Usually, problems that ask you to calculate a substance's specific heat will provide such information (heat, `DeltaT`, mass).