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November 26, 2014

November 26, 2014

Posted by **Phillip** on Sunday, April 24, 2011 at 5:50pm.

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

- Chemistry -
**DrBob222**, Sunday, April 24, 2011 at 6:15pmYou 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.

- Chemistry -
**Phillip**, Sunday, April 24, 2011 at 6:23pmI'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?

- Chemistry -
**DrBob222**, Sunday, April 24, 2011 at 6:45pmYes. 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.

- Chemistry -
**Phillip**, Sunday, April 24, 2011 at 7:05pmThank you so much DrBob!

I ended up with

469.5Tf = 8122.54

Tf = 17.3

Thanks again,

-Phil

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