Homework Help: Specific heat?

1. Dec 6, 2003

tucky

specific heat????

Hey guys….I need help with another problem.

Q: Below are two sets of data from an experiment to determine the specific heat of a metal sample. For which mass will t the uncertainty n the specific heat be greater? Explain your answer. The uncertainty in all temperature measurements is +/- 1 degree C. And for all masses are +/- .1g. The specific heat of water is 1.0 cal/g degree C with negligible uncertainty.

Metal Sample A:
Initial temperature A----25 C
Mass of metal A-------- 100g
Initial Temp of water----100 C
Mass of water-----200g
Thermal Equilibrium temp of metal B in water---75 C

Metal Sample B:
Initial temp B----25C
Mass of metal----100g
Initial temp of water----100C
Mass of water----200g
Thermal Equilibrium temp of metal B in water---90C

A: Q=m*C*(change temp)

Metal A Q=(.1 +/-.0001kg)C(348+/-1K-298+/-1k)
Metal A Q=5+/-.19716KgKC

Metal B Q=(.1 +/-.0001kg)C(363+/-1K-298+/-1k)
Metal B Q=6.5+/-.29705KgKC

Now, I am stuck…I don’t know how to finish this problem. Can anyone help me??? Actually, I don’t even know if I am on the right track.

Tucky

2. Dec 7, 2003

arcnets

I think this is an exercise in error propagation.
If you have a function
$$f = f(x, y, z),$$
then the uncertainty is
$$df = \sqrt{ (\frac{\partial f}{\partial x}dx)^2 + (\frac{\partial f}{\partial y}dy)^2 + (\frac{\partial f}{\partial z}dz)^2}.$$

3. Dec 7, 2003

tucky

Is there a way to do work this problem with out calculus? Because my class is an algebra/trig based class. Thank you for your help.

4. Dec 7, 2003

arcnets

Well, in the 2 experiments all data are the same, except for the final temperature. So you could argue that the larger change in temperature corresponds to the smaller relative error.