Predicting Heat Loss from Glass Beaker Experiment

AI Thread Summary
The discussion revolves around an experiment measuring heat loss from a glass beaker filled with water. The user calculated thermal current using the equation I = kA (ΔT/Δx) but expressed concern that their results seemed excessively high. They initially calculated the cross-sectional area of the beaker incorrectly and later realized they needed to account for the sides of the beaker as well. Participants suggested that the user verify their calculations and units, emphasizing that the results should be more reasonable. The user seeks guidance on converting watts to a temperature loss rate for better comparison with experimental data.
vr_ben
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Homework Statement


Alright so I am doing a fairly simple experiment testing how the temperature of water in a beaker effects how fast heat is lost. Have done the tests, but need to get some predicted results to compare them too.

Have been using a glass beaker filled with 200ml of water, beaker dimensions being 60mm diametre, 150mm tall, about 1.5mm thick (actually forgot to measure this, so if this is off let me know :P)

Thermal conductivity of glass: 0.8

The Attempt at a Solution


Ok so i used an equation for thermal current:
I = kA (Change in T/ Chane in x) with I being current (Watts) k being thermal conductivity, A being cross sectional area (i calcualted 28.27cm^2 but not sure if that's correct), T being temperature (Kelvin), x as distance current travels.

Anyway using that equation i calcualted it for a few different temperatures and got:
change in temp - Thermal current
70 - 10554.13W
60 - 9646.4W
50 - 7538.6W
40 - 6030.93W
30 - 4323.2W
20 - 3015.46W

And this is where I am sort of stuck. First of all are those reasonable numbers (seem quite high to me...)? and is there a way to convert W to degrees per second or minute or something which i could actually compare to real results?

Anyway if you read all that and can help in anyway way its much appreciated
thanks
 
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umm bump? anyone able to help, am i even using the right formula?
 
How did you get that area? It is too small.

ehild
 
oh i think that was just the area of the bottom, maybe, did this a while ago.. suppose i need to include the sides aswell? Didnt get real measurements of the beaker which is a shame... but including the side it would be something like 250cm^2?

That makes the current even higher.

Anyway thanks for that, is there a way to get a temperature per second loss or something from the answers in W? (is a watt = JS^-1?, and if so can i change that to temperature loss?)
 
Your formula for the heat loss per unit time which is Js^-1 is correct, but the results are really too high. Check the units and your calculations.
As for the area of the side of the baker that conducts heat, take into account that the 200 ml water does not fill the baker.

ehild
 
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