Question

Calculate the calorimeter constant if 25 g of water at 54°C was added to 25 g of water at 25°C with a resulting temperature of 35°C

Answer 1

We need to find the difference between the heat lost by the hot water when it droped from 60.0 to 40.0 and the heat gained by the cold water when it was heated up to 40.0 from 25.0. Note that the values in the problem are in mL and the values in the solution are grams. The volume (mL) is converted to the mass (grams) by using the density of water (1.00 g/mL).

1) Hot water lost:

q = m Δt Cp

q = (40.0 g) (20.0 °C) (4.184 J g¯1 °C¯1)

q = 3347.2 J

2) Cold water got:

q = m Δt Cp

q = (40.0 g) (15.0 °C) (4.184 J g¯1 °C¯1)

q = 2510.4 J

3) The calorimeter got the rest:

3347.2 − 2510.4 = 836.8 J

4) Find the heat capacity of the calorimeter:

836.8 J / 15.0 °C = 55.8 J / °C