A piece of metal of mass 12 g at 104◦C is

Question

placed in a calorimeter containing 45.4 g of
water at 22◦C. The final temperature of the
mixture is 71.5
◦C. What is the specific heat
capacity of the metal? Assume that there is
no energy lost to the surroundings.
Answer in units of J
g ·
◦ C

Answer 1

The sum of the heats gained is zero.

heatgainedmetal+heatgained water=0
12*Cm*(Tf-Tim)+45.4*Cw*(Tf-Tiw)=0

Tf=71.5 Tim=104 Tiw=22
solve for cm

Answer 2

I plug in the numbers and solve using a calculator, this is how i did it.

12(71.5-104)+45(71.5-22)=1857.3 J/g C

I still get the answer wrong.

Any errors i am making?

Answer 3

Sorry for posting twice :(

Answer 4

of course you get the wrong answer. What happened to cm,cw in my equation. You are looking for cm. Do some algebra.

12*Cm*(Tf-Tim)+45.4*Cw*(Tf-Tiw)=0

Tf=71.5 Tim=104 Tiw=22
solve for cm