Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Predicting heat loss

  1. May 28, 2010 #1
    1. The problem statement, all variables and given/known data
    Alright so im 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

    3. 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 thats 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 im 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
  2. jcsd
  3. May 29, 2010 #2
    umm bump? anyone able to help, am i even using the right formula?
  4. May 30, 2010 #3


    User Avatar
    Homework Helper

    How did you get that area? It is too small.

  5. May 30, 2010 #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?)
  6. May 30, 2010 #5


    User Avatar
    Homework Helper

    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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook