Posted by **Kate** on Tuesday, October 28, 2008 at 4:54pm.

I need help!

I found the change in Temp= -.9 degree C. I don't know what to do next...?

What value from this problem do I use as my q(heat capacity) to plug into the equation q= m X Cs X change T?

The question states:

120mL of H2O is intially @ room temp (22 degree C). A chilled steep rod @ 2 degree C is placed in the water. If the final temp of the system is 21.1 degree C, what is the mass of the rod?

Thank You!!!

- Chemistry -
**DrBob222**, Tuesday, October 28, 2008 at 5:07pm
The rod is cooler than the water; therefore, the rod will gain heat and the water will lose heat and the two added together will be zero (one will be negative).

massrod x specific heat rod x (Tfinal-Tinitial) + massH2O x specific heat water x (Tfinal-Tinitial) = 0

Solve for

You will need to look up the specific heat of the rod and the water if they are not listed in the problem. Post your work if you get stuck.

- Chemistry -
**DrBob222**, Tuesday, October 28, 2008 at 6:15pm
I think trying to shuffle the equation with algebra makes the problem harder. I think what you did wrong is that you didn't substitute Tf-Ti correctly.

Here is what I did:

massrod x sp.h.rod x (Tf-Ti) + massH2O x sph.H2O x (Tf-Ti) = 0

mass rod = Y

[Y*0.452*(21.1-2)] + [120*4.18*(21.1-22.0)] = 0

8.633Y-451.44 = 0

PLEASE check what I've done to make sure I've not made an error. You can finish the problem but the answer I obtained is approximately 50 g. It appears to me that you substituted 1 for Tf-Ti for both rod and water. I hope this helps. Let me know if you have any problems.

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