Posted by Kate on Tuesday, October 28, 2008 at 4:54pm.
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.
I did not get the correct answer...
Solve for MassSteel.
Here is my work-
masssteelx specific heat rod x (Tfinal-Tinitial) + massH2O x specific heat water x (Tfinal-Tinitial) = 0
MassSteel = -massH2O x specific heat water x (Tfinal-Tinitial)/ specific heat rod x (Tfinal-Tinitial)
MassSteel = -120mL x 4.18J/gC x -1C/ 0.452 J/gC x -1C
MassSteel = 1109.73
What did I do wrong???
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.
It worked!! Thanks so much for your help!