[sara]

please show me what I'm doing wrong!!
A calorimeter contains 75.0 g. of H2O at 16.95degree Celcius. 93.3 g. of Fe was placed in it, giving a final temp. of 19.68 Celcius.Calculate heat capacity of the calorimeter. Specific heats are: 4.184 J/g.degree Celcius for water & 0.444 J/g. Celcius for Fe.
Mass Fe= 93.3 g
^ Temp. Fe= 65.58 C - 19.68 C = 45.90 C
M. H2O= 75.0g
SH water= 4.184 J/g. C
^ Temp. H2O= 19.68 C-16.95 C=2.73C
^ Temp. cal.= 19.68C-16.95C=2.73C

HCc= (93.3 X 0.444 X 45.90C) - (75.0 X 4.184 X 2.73) divided by 2.73 C = 2.44 J/C.

should be" 381 J/ C  HOW !!!!!

[dexangeles]

first, your change in temperature on your Fe is wrong it's final - initial
but you have seem to reach the right equation in the end
i assume you already know how to set up the calorimetry equation

check your calculation again cause it gives 382.7 J/C
which is the same as 381 J/C
there will be round off errors due to significant figures

93.3g x 0.444J/g C x 45.9C = 1901.41668J or can be rounded off to 1900J
75.0g x 4.184 J/g C x 2.73C = 856.674J or can be rounded off to 860J

(1900J - 860J)/2.73C = 380.95J/C or 381 J/C

don't ask me why they round off within the calculations
they should know it causes round off errors