Question

the following experiment was conducted to determine the heat capacity of a calorimeter. the mass of the empty calorimeter was determined. A quantity of cold water was added to the calorimeter after whih both the temperature and new mass were determined. A second quantity of water was heated and its temperature was determined. The heated water was added to the cold water in the calorimeter after which both the temperature and mass of the calorimeter plus water were determined. All together three mass measurements and three temperature measurements were made as indicated below. Determine.

Answer 1

The question's incomplete. what are the measurements?

Answer 2

determine the heat capacity (in J/C) of the calorimeter from these data. The specific heat capacity of water is 4.18 J/gC.

Mass of calorimeter: 11.33g
Mass of calorimeter + cold water: 57.33g
Temp of cold water: 23.6 degree C
Temp of heated water: 33.1 degree C
mass of calorimeter _ cold water + heated water: 102.33g
temp of mixed cold water and heated water: 28.2 degree C

Answer 3

mass of cold water=57.33-11.33=46g.
Mass of hot water=102.33-11.33=91g.
Initial temp of cold water and calorimeter=23.6 and hot water=33.1
Final tempertaure:28.2
I assume you know the equation Q=MC deltaT
Heat required to raise water from 23.6 to 28.2= 46*4.18*4.6.
Heat required for calorimeter=11.33*C*4.6
Heat released by hot water = 91*4.18*4.9
Ok till here?

Answer 4

Now by calorimetric law,
46*4.18*4.6+11.33*c*4.6=91.4*4.18*4.9.
Solve for C to get the answer.

Answer 5

thank you

Answer 6

no problem :)