Question

40.0 grams of water at 64.2C is added to 60.0 grams of water at 25.0C in a calorimeter. The final temperature of this solution was found to be 39.6C. What is the heat capacity of the calorimeter?

Answer 1

Specific heat of water= 4.184 J/gm.deg.c

heat is removed from water at 64.2 deg.c and this is added to water at 25 deg.c and calorimeter

hence heat removed from hot water= mass of water* specific heat of water* change in temperature = 40*4.184*(64.2-39.6) =4117 = mass of cold water* specific heat of cold water* change in temperature + calorimeter constant* change in temperature = 60*4.184*(39.6-25)+Calorimeter constant*(39.6-25)

Calorimeter constant = heat capacity of calorimeter = (4117-3665)/ (39.6-25)= 30.95 J/deg.c