Question

A 150.0g piece of metal (c = 0.560 J/g•C) is dropped into a calorimeter which contains 40.0g of water. The metal was initially at 55°C and the calorimeter was initially at 25°C. The mass of the calorimeter is 5.0 g. If the temperature of the calorimeter increased to 29°C what is the c of the calorimeter?

Answer 1

for metal : Q = m Cp dT

                         = 150 x 0.560 x (55 -29)

                        = 2184 J

for water : Q = m Cp dT

                         = 40 x 4.184 x (29-25)

                         = 669.44 J

dT = 29-25 = 4

heat lost by metal = heat gain by calorimeter

2184 = 669.44   + C x 4

C = 378.6 J / oC

calorimeter constant = C = 379 J / ^oC