posted by Kate on .
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?
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
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!