Predicting Heat Loss from Glass Beaker Experiment

In summary, the conversation is about an experiment testing the effects of water temperature on heat loss in a beaker. The person conducting the experiment used an equation for thermal current and calculated the results for various temperatures. However, there were concerns about the accuracy of the results and whether the formula was being used correctly. It was also mentioned that the area used in the calculations may be too small and should include the sides of the beaker. Further clarification and assistance were requested.
  • #1
vr_ben
3
0

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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  • #2
umm bump? anyone able to help, am i even using the right formula?
 
  • #3
How did you get that area? It is too small.

ehild
 
  • #4
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?)
 
  • #5
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
 

1. How do you conduct a heat loss experiment using a glass beaker?

The experiment involves filling a glass beaker with a known amount of hot water and measuring its temperature at regular intervals. The rate of temperature change can be used to calculate the heat loss from the beaker.

2. What factors can affect the heat loss from a glass beaker?

The surface area of the beaker, the initial temperature of the water, the temperature of the surrounding environment, and the material of the beaker can all affect the rate of heat loss.

3. How do you calculate the heat loss from a glass beaker experiment?

The heat loss can be calculated using the formula Q = mcΔT, where Q is the heat lost, m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature over time.

4. Can the heat loss from a glass beaker experiment be used to predict real-world scenarios?

Yes, the results from a glass beaker experiment can be used to make predictions about heat loss in real-world scenarios, such as in buildings or industrial processes.

5. How can the results of a glass beaker heat loss experiment be improved?

The experiment can be improved by conducting multiple trials, using a larger beaker to reduce the surface area to volume ratio, and controlling for variables such as room temperature and water temperature. Additionally, using a more accurate temperature measuring device can improve the accuracy of the results.

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