How Does Mechanical Energy Convert to Heat in Lead Shot Experiments?

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Homework Help Overview

The discussion revolves around the conversion of mechanical energy to heat in a lead shot experiment. The original poster describes a setup involving a tube with lead shot that falls a certain distance, and they are attempting to calculate the mechanical equivalent of heat generated during the process.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between potential energy, kinetic energy, and heat generation, questioning whether the heat calculated from mechanical energy should equal the heat calculated from temperature change. Some participants raise concerns about energy losses due to sound and other factors.

Discussion Status

There is an ongoing exploration of the discrepancies between theoretical calculations and experimental results. Some participants suggest that the energy lost to sound and other interactions may account for differences in expected outcomes. Others propose alternative experimental setups to isolate variables affecting temperature change.

Contextual Notes

Participants note the complexity of real-world experiments compared to theoretical calculations, highlighting the influence of various factors that may not be accounted for in simplified models. There is also mention of the need for careful consideration of assumptions regarding energy conversion in the experiment.

Geoff
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Hello.
I need to calc the mechanical equivalent of heat.
I got a tube with some lead shot in it.
Tube is 1 m long. Lead shot is 11.3 kg.
It falls though 66 cm of the tube. (66 cm of the tube are empty).
It was fliped 100 times
temperature rose 10 degrees.
SO we get:
Q=11.3*130*10
where 130 is the specific heat of lead.
We got some number there.
That is heat.
Now we do:
Joules of mechanical energy=11.3*0.66*9.8*100(number of times flopped)
Now we get another number.

The two number need to be close to each other right? Or am i missing something when i am relating mechanical energy with heat?
the heat that was generated from Q=mcdeltat should equal the mgh#oftimesflipped, right?
Sry for the dumb post, but i don;t have a textbook on hand and i haven;t done this in a long while...
Thx
 
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the energy you have given there is potential energy of a falling object. the mechanical energy is the total energy which is KE+V. the loss of kinetic energy is usually converted to heat or sound, so the energy due to heat will be less than the potential energy.

I think I am right in saying this
 
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y would it be less?
Cuz the way heat is generated is when the lead falls the 66 cm shaft and hits the bottom, it generates heat. So the change from mgh to Ke to Heat energy shoudl all be equivalent. THere will be major loses due to sound and because the system is not adiabatic, but under ideal conditions should the q=mct and mgh*#offlips should be the same rite?
Thx
 
Eleven kilograms is quite a lot. Don't you mean 11 grams?
 
The kinetic energy gained by the lead shot is going to be less than the potential energy because you lose some heat to sound.
 
As you swing the lead from the bottom to the top you give it some kinetic energy. When you stop it at the top this energy needs to go somewhere (it does not simply vanish into thin air). You did work on the lead shot to give it this kinetic energy, it is not a result of a decrease in potential energy. Quantifying this change in kinetic energy will be challenging though.
 
Doing experiments is quite a different process than doing calculations in Physics. When we do calculations we ignore details in order to get to the answer. In real life this is not possible. All effects are in operation and will influence the result. This makes it difficult to get to the bottom of things. This means that you are on your own when you do a experiment. Unless you can find someone who has done a similar experiment you have to draw your own conclusions. Having said that I confess that I have not done the experiment myself so these are suggestions as to what may be the cause of the discrepencancy in your results. It is up to you to investigte their validity or reject them on a theoretical basis.

The first attachment shows some interaction forces arising on the lead shot as they hit the bottom of the tube. It is these interactions that cause the lead shot to heat up.

The second attachment shows the same sort of interaction forces arising as the tube is turned sideways.

In the third attachment one sees the same situation arising as the turning motion is stopped at the top.

Could it be that an acceleration of the lead shot heats it up? If so it is experiences an acceleration all the way up due to a required centripetal force. Admittedly it is much smaller, but it acts over a longer time.

One can envisage an experiment where this turning effect is eliminated with maybe a tube with a platform at the bottom and an attached plunger sticking out. The platform can then be raised with the lead on it and quickly lowered. The same interacting forces will arise when the lead shot is accelerated initially. This suggests that one should raise them very slowly in order to minimize the magnitude of this effect. So we conclude that it is not just the falling down that caused the lead shot to gain heat, but also the raising. Could this maybe explain the resulting temperature to be double what one would expect?
 
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I am not to happy to report the result of an investigation that I made regarding the increase in temperature of lead shot due to the conversion of mechanical energy. I mounted a bicycle wheel horizontally and checked it with a radially mounted spirit level. Next a bottle with 308 grams of lead shot was mounted to the rim at a distance of (35.0 +/- 1.0) cm from the axle and rotated from the one side to the other halfway along the perimeter 100 times. A temperature increase of only 0.7 degrees celsius was observed. My previous hypothesis that the rotation of the tube might be responsible for the additional temperature increase therefore seems to be wrong since the expected temperature increase due to the conversion of the potential energy should be
[tex]\Delta T = \frac{9.79 \times 70}{128}\ =\ 5.4\ C^o[/tex]
in this case.
 

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