Help with chemistry problem please!!

Question

If 100 ml of water at 72 C is mixed into a calorimetry apparatus that has 125 ml of water at

18.2 O C, and the final temperature of the mixture is 41 C, then what is the calibration constant

(C) of the calorimetry apparatus?

Answer 1

heat lost by the warm water must equal heat gained by the cooler water PLUS the heat gained by the calorimeter (calorimeter constant). In equation form...

heat lost by warm water = q = mC∆T

q = heat

m = mass of water = 100 g

C = specific heat of water = 4.184 J/gº

∆T = change in temperature = 31º

q = (100 g)(4.184 J/gº)(31º)

q = 12,970 J of heat lost by the warm water

This must be the heat gained by the cold water + the heat gained by the calorimeter

12,970 J = mass of cool water x specific heat x change in temp + Ccal∆T where Ccal is the calorimeter const.

12,970 J = (125 g)(4.184 J/gº)(22.8º) + 22.8C_cal

12,970 J = 11,924 J + 22.8C_cal

C_cal = 45.9 J/º