Given the following information, calculate the heat capacity of the calorimeter in the unit of J/ degree Celsius: A calorimeter contains 19.16 g of water, both temperatures are at 22.5 degree Celsius. 14.82 g of water with temperature at 68.4 degree Celsius is poured in, and stirred. The final temperature of all components is 40.1 degree Celsius.
-
heat lost by hot water = mass x c x ∆T
heat = 14.82g x 4.184J/g-ºC x 28.3ºC = 1754.79J
1754.79J = ∆Tcold x (mass water x specific heat water + heat capcity cal)
1754.79J = 17.6ºC x (19.16g x 4.184J-gºC + Ccal)
1754.79J = 1410.91J + 17.6xC
343.87 / 17.6 = C
19.54J/ºC
heat = 14.82g x 4.184J/g-ºC x 28.3ºC = 1754.79J
1754.79J = ∆Tcold x (mass water x specific heat water + heat capcity cal)
1754.79J = 17.6ºC x (19.16g x 4.184J-gºC + Ccal)
1754.79J = 1410.91J + 17.6xC
343.87 / 17.6 = C
19.54J/ºC