Lab 10 Preparation

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Lab 10 Preparation

Post  Kathleen on Mon Dec 08, 2008 9:50 pm

This is pretty straightforward but I'll post it anyway
1.) theta = N*2*pi
torque = F*R
W= tau*theta
W=(F*R)*(N*2*pi)

2.) Q = m*c*deltaT
the mass of the object = m
the specific heat " = c
and the temperature rise " = deltaT
The combined heat for all objects is Qi + Qo + Qw + Qs + Qt

3.)
A- Crank turning the black cylinder
B- Counter for the # of turns
C-String attached to the disk which is attached to the inner brass cup
D-Spring scale measuring the tension in the string
E- Outer brass cup attached to the turning black cylinder
F- Inner brass cup, filled with water and attached to the disk held by the string
G-Rotating black cylinder attached to the outer brass cup
I-Thermometer reaching into the water
J- Stirrer to distribute the warm water for equilibrium

4.) The following statements are correct:
-Suppose you have the same mass for all materials in the right sketch above, then the amounts of heat absorbed by the materials can be ordered as: Qwater>Qthermometer>Qbrass
-If one starts the experiment with the water at room temperature the amount of heat using equation (2) for the objects listed in the right figure above is underestimated.
-When one starts the water out at ~ 5 degrees C below room temperature and ends at a water temperature of ~ 5 degrees C above room temperature the amount of heat gained from the environment is ~ equal to the amount of heat lost to it.

Kathleen
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Re: Lab 10 Preparation

Post  xxcutiex on Thu Dec 11, 2008 5:04 pm

thank you very much Surprised) appreciate the help

xxcutiex
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