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Thermal conductivity of stainless steel

  1. Apr 17, 2003 #1

    xyz

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    Hi...
    there is a question i can't figure out. "A 400-g stainless steel tea kettle containing 500g of water is on top of the stove.The portion of the tea kettle that is in contact with the heating element has an area of 0.005m^2 and is 2.5mm thick. At a certain moment,the temperature of the water is 75C,and it is rising at the rate of 4C per minute. What is the difference in temperature between the inside and the outside of the bottom of the tea kettle? the thermal conductivity of stainless steel is 16.3W/(m K) and the specific heat is 448 J/(kg K)."
    Thank you for any replies!!!
     
  2. jcsd
  3. Apr 17, 2003 #2
    Well.
    Working on the laws of this forum, we are not able to give you ready answers, you should tell us where you got stuck !
    I will give you some clues ...
    Knowing that the temprature rises in a certain ammount in each time unit, and knowing the specific heat of water (and the mass of water), you can calculate the ammount of energy that is reaching the water in time unit (or the heat power on water).

    Now this is supposed to equal the heat power giong through the kettle (since the heat is coming from it to the water, and assuming there is no heat lost), if you equal both of them, you should figure out [del]T.

    If you tell us where you got stuck, we may be more usefull :smile:
     
  4. Apr 17, 2003 #3

    xyz

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    humm...thanks:wink:
    I am just confused the relationship of heat conduction,convection and radiation, like the heat flows from the stove,then the heat flow into the tea kettle,it causes different temperatures between the tea kettle and water.

    In this case,i have to compute the Temperature of the tea kettle rising per minute.Right??
     
  5. Apr 17, 2003 #4

    xyz

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    Hi...STAii

    i calculate the amount of energy that is reaching the water in 1 minute, m c [del] T=0.5kg*4186j/(kg K)*4C=8372j.

    8372j equals to the heat power going through the kettle,
    8372j=k A [del]T *60s/L =16.3W/(m K)*0.005m^2* [del]T*60s/0.0025
    so i compute [del]T=4.28C...the difference in temperature between the inside and the outside of the bottom of the kettle is 4.28C...

    Am i right? looking forward ur reply...
     
  6. Apr 18, 2003 #5
    The concept is right (i didn't look into the numbers ..), but there are some little notes.

    "8372j equals to the heat power going through the kettle"
    Joule is not a unit of power, it is a unit of energy, so you either have to say :
    "8372 j/min equals to the heat power going through the kettle"
    or say:
    "8372j equals to the heat eneryg going through the kettle in a single minute"


    Otherwise, everything is fine.
    Also note that it is convenient to use SI units, like you could have used the unit of second for time, and therefore found the heat power reaching the kettle in Watts, and used the following equation to solve.
    P(ower In Conductivity) = (k*A*[del]T)/L

    Hope i helped.
     
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