Question

Find the q_water, q_calorimeter, q_metal(unknown) with the following data.

Note: q_{water +} q_{calorimeter +} q_metal = 0

Mass of empty calorimeter: 161.61 g

Mass of Calorimeter + water: 201.03 g

Mass of Calorimeter + water + metal pieces: 216.04 g

Initial Temp: 27.214 ^oC [Calorimeter + water + metal pieces]

Final Temp: 31.975 ^oC [Calorimeter + water + metal pieces]

Initial temp of metal pieces: 100 ^oC

Answer 1

-Qmetal = Qwater + Qcal

-mmetal * Cpmetal (Tf-Tmetal) = mwater*Cpwater*(Tf-Tw) + CCal(Tf-Tw)

mmetal = 216.04 - 201.03 = 15.01 g

mwater = 201.03 - 161.61 = 39.42

-15.01 * Cpmetal (31.975-100) = 39.42*4.184*(31.975-27.214 ) + CCal(31.975-27.214 )

Since no data on Ccal (heat cpaacity of calorimter ) is given, then assume CCal = 0

-15.01 * Cpmetal (31.975-100) = 39.42*4.184*(31.975-27.214 )

Cpmetal = 39.42*4.184*(31.975-27.214 ) / (-15.01 * (31.975-100))

Cpmetal = 0.769J/gC

Qmetal = -15.01 * 0.769 * (31.975-100) = -785.19 J

Qwater + Qcal = -Qmetal = 785.19 J