Determination of the Calorimeter Constant
When 31.152 grams of water at 51.0 °C are mixed with 31.284 grams of water in a calorimeter at 20.2 °C, theequilibrium temperature attained by the system is 34.8 °C . The specific heat of water is 4.18 J / g°C . Calculate the calorimeter constant for this system in J/°C.
Calibration of StyrofoamCup Calorimeter
Figure 1 Thermogram for the addition of 31.152 grams of water at 51.0 °C to31.284 grams of water at 20.2 °C . The equilibrium temperature of the system is 34.8°C. D T of the warm water is 16.2 °c, and DT of the cool water is 14.6°C.
-qwarm water = qcool water +qcalorimeter
qcalorimeter = -qwarm water - qcool water
qcalorimeter = -(mwarm water X Cwarm water X DTwarm water) - (mcool water X Ccoo1 water X Dcool water)
qcalorimeter =-[31.152 g X 4.18 J X (34.8 - 51.0) °C]_ [31.284 g X 4.18 J X (34.8 - 20.2) °C]
g°C g °C
qcalorimeter = 2110 J - 1910 J
NOTE 1: DT calorimeter equals DTcool water because thecalorimeter is initially at the same temperature as the cool water and both absorb energy from the warm substance.
qcalorimeter = 200. J
Ccalorimeter = qcalorimeter/DTcool water [NOTEI)
C = 200.J =13.7J°/C
calorrmeter (34.8 °C – 20.2°C .The calorimeter constant for the styrofoam calorimeter indicates that the calorimeter absorbs 13.7 J of heat for every 1.00 °C increase in the temperature of the system. Thecalorimeter constant should have a small, positive value. The value calculated above will be used in the following sample calculation to determine the specific heat of an unknown substance.