1. The problem statement, all variables and given/known data 50ml of water at 49.6 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. 2. Relevant equations Density of water = 1.00 g/mL Specific heat capacity = 4.18 J / g * K 3. The attempt at a solution q (heat given up by water) = 50ml * (49.6-30.1) = 840 cal q (heat absorbed by cold water) = 50ml (30.1-25.1) = 250 cal Heat absorbed by calorimeter = 250 + 840 = 1090 cal Ccal = qcal / delta T = 590 / (30.1-25.1) = 118 K The answer is supposed to be 493.24 J/K .... but i am not getting that. Please someone please show me how to correct this. Thank you.