Which Metal Sample Shows Greater Uncertainty in Specific Heat Calculation?

AI Thread Summary
The discussion centers on determining which metal sample, A or B, exhibits greater uncertainty in specific heat calculations based on given experimental data. Both samples have the same mass and initial conditions, but differ in their thermal equilibrium temperatures. The calculations show that Metal A has a specific heat uncertainty of 5 +/- 0.19716 KgK°C, while Metal B has a higher uncertainty of 6.5 +/- 0.29705 KgK°C. The conversation highlights that the larger temperature change in Metal B may lead to a smaller relative error, suggesting that the uncertainty is influenced by the temperature difference. The problem emphasizes the importance of error propagation in experimental physics, particularly in algebra-based contexts.
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
<br /> f = f(x, y, z),<br />
then the uncertainty is
<br /> df = \sqrt{ (\frac{\partial f}{\partial x}dx)^2 + (\frac{\partial f}{\partial y}dy)^2 + (\frac{\partial f}{\partial z}dz)^2}.<br />
 
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top