Question

A certain calorimeter absorbs 65 J/oC. If 55.0 g of 67oC water is mixed with the calorimeter’s original 50.0 g of 23oC water. Calculate for the final temperature of the mixture?

Answer 1

Q = C x m x deltaT

C_water = 4.186 J/g°C

Q1 = 4.186 x 55 x 67 = 15 425 J

Q2 = 4.186 x 50 x 23 = 4 814 J

Q3 = 15 425 + 4 814 = 20 239 J

t = 20 239 / ((50 + 55) x 4.186) = 46.1°C