If 150.0 g of copper at 100.0 C and 125.0g of liquid water at 35.0 C are mixed in an insulated container, what is the final temperature of the mixture? Ccu= .39 J/g C Cwater= 4.184 J/g C
Did i do this correct:
150.0 x 100.0 x .39 = 250 J
125.0 x 35.0 x 4.184 = 18300 J
18300 + 250= 18550 J
I used sig fig.
2 answers
The first multiplication should be higher
i did
100j/.39x150.0g=1.7c use sig fig so its 2
then you subtract from the H2O the temp. change
100j/.39x150.0g=1.7c use sig fig so its 2
then you subtract from the H2O the temp. change