A piece of iron weighing 62.3g was heated to 129*C and then dropped into a container containing 209.4g of water at 23.1*C. What is the final temp of the water when thermal equilibrium is reached? C(H2O)=4.184J/gK C(Fe)=0.449J/gK

Question

A piece of iron weighing 62.3g was heated to 129*C and then dropped into a container containing 209.4g of water at 23.1*C. What is the final temp of the water when thermal equilibrium is reached? C(H2O)=4.184J/gK  C(Fe)=0.449J/gK

anonymous · Answer

I know I am using q=cm (delta)T and that q(Fe)+q(H20)=0

anonymous · Answer

Just not sure where to plug everything in.

anonymous · Answer

I keep getting 3-6*C off from the answer which I know is 26.4*C or 299.6K

aaronq · Answer

equate them to each other, 

deltaT= Tf- Ti

    iron               water
mC(Tf- Ti) = mC(Tf- Ti)

anonymous · Answer

So I would set it up like  (0.449)(62.4)(x-129.2)=(4.184)(209.4)(23.1)

anonymous · Answer

I am getting 19.6*C

anonymous · Answer

My teacher mentioned in her lecture that qFe=-qH2O

anonymous · Answer

And that qFe=-CH20m(tf-ti) But I am not getting the right answer it that either.

aaronq · Answer

sorry i did forget the - sign

 (0.449)(62.4)(x-129.2)=- (4.184)(209.4)(x-23.1)

aaronq · Answer

that was close, but you didn't add the x on the water side

anonymous · Answer

Oh ok. So C for water is negative but C for iron stays positive. doing it that way gets me the right answer. Thank you so much. This is a study problem for a test we have on Monday and I just couldn't figure it out. Thanks again.

aaronq · Answer

no problem. well it's not the C that is negative

it's because the heat, q, given off by the iron is fully absorbed by the water

so yeah  $$q _{Fe}=-q _{H _{2}0}$$