Question

50ml of water at 46.9 C were mixed with 50ml of water at 25.1 C in a calorimeter also at 25.1 C. The final temperature was 30.1 C Assuming that neither the density of water nor its specific heat capacity change with temperature, calculate the total heat capacity of the calorimeter.
Density of water = 1.00 g/mL
Specific heat capacity = 4.18 J / g * K

Answer 1

heat lost by hot water + heat gained by cold water + heat gained by calorimeter = 0
[mass hot water x specific heat water x (Tfinal - Tinitial)] + [mass cold water x specific heat water x (Tf-Ti)] + Cp(Tf-Ti) = 0.
Solve for Cp.

Answer 2

Cp = 0.49 J * K^-1