Which Metal Sample Shows Greater Uncertainty in Specific Heat Calculation?

Click For Summary

Homework Help Overview

The discussion revolves around determining which of two metal samples exhibits greater uncertainty in specific heat calculations based on experimental data. The context involves understanding error propagation and the effects of temperature changes on uncertainty in measurements.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the concept of error propagation and its application to the problem. Some question whether calculus is necessary for the solution, while others discuss the relationship between temperature change and relative error.

Discussion Status

The discussion is ongoing, with participants offering insights into error propagation and questioning the need for calculus. There is an exploration of how the final temperature affects the uncertainty in specific heat calculations.

Contextual Notes

Participants note that all temperature measurements have a fixed uncertainty of +/- 1 degree C, and mass measurements have an uncertainty of +/- 0.1g. The specific heat of water is considered to have negligible uncertainty.

tucky
Messages
30
Reaction score
0
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
 
Physics news on Phys.org
I think this is an exercise in error propagation.
If you have a function
[tex] f = f(x, y, z),[/tex]
then the uncertainty is
[tex] df = \sqrt{ (\frac{\partial f}{\partial x}dx)^2 + (\frac{\partial f}{\partial y}dy)^2 + (\frac{\partial f}{\partial z}dz)^2}.[/tex]
 
Is there a way to do work this problem without calculus? Because my class is an algebra/trig based class. Thank you for your help.
 
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.
 

Similar threads

Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
54
Views
10K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
2
Views
11K
Replies
2
Views
2K