A 32.5g cube of aluminum initially at 45.8 C is submerged into 105.3g water at 15.4 C. What is the final temp of both substances at thermal equilibrium. Assume the al and the H2O are thermally isolated from everything else)

Question

Specific heat cap Al= .903
Specific heat cap H2O= 4.18
I understand this problem but In my math i get to:
(DeltaT) Al = -14.998 X (DeltaT) H2O
And im lost. Anyone able to explain the completion of the problem? Any help is greatly appreciated

Answer 1

You need to use Tfinal-Tinitial for delta T.
heat lost by Al + heat gained by water=0
[mass Al x specific heat Al x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tf.

Answer 2

I'm having trouble once i get to

T(final)=-14.998 X T(final) + 14.998 X T(initial h2o) + T(initial Al)

Any suggestions?

Answer 3

Yes. I think it is tough to try to manipulate the algebra. It is much easier to substitute the numbers first and manipulate them.
[32.5 x 0.903 x (Tf-45.8)]+[105.3 x 4.18 x (Tf-15.4)] = 0
29.35Tf - 1344.1 + 440.15Tf - 6778.4 = 0
Check those numbers to make sure I didn't make an error on my calculator, then solve for Tf.

Answer 4

Thank you so much DrBob!

I ended up with

469.5Tf = 8122.54
Tf = 17.3

Thanks again,
-Phil