A jug containing 0.250 kg of liquid is put into a refrigerator. Its temperature decreased from 20°C to 15°C. The amount of energy transferred from the liquid was 5,250 J. Calculate the specific heat capacity of the liquid.

Using the equation Q = mcT, where Q represents energy in Joules, m represents mass in kg, c represents specific heat capacity measure in joules kg^-1 K^-1 and lastly, T represents the change in temperature in celcius or kelvin. Firstly rearrange the equation to make specific heat capacity the subject of the equation. This will result in the equation c = Q/(mT). Now we can use the figures from the explanation of the problem and replace the variables in the equation.
We are told that the mass of the liquid is 0.250 kg, the change in temperature is from 20 degrees celcius to 15 degrees celcius so there is a change in 5 degrees celsius and the total energy transferred was 5250 J. So by replacing the variables with the given figures we get the result that c = (5250)/(0.250 x 5), by simplifying this we get the result that the specific heat capacity of the liquid is 4200 J kg^-1 K^-1.