If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter increases from 24.58°C to 35.82°C. Calculate the heat capacity of the calorimeter (in J/°C). The specific heat of water is 4.184

Answer 1

q = Ccal*(Tfinal-Tinitial) + [mass H2O x specific heat H2O x (Tfinal-Tinitial)]
Solve for Ccal
Watch the units. If you use 4.184 J/g*C for H2O and you want Ccal in J/C, then you must substitute q in J, not kJ.