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Newton Law of Cooling

  1. Dec 7, 2011 #1
    A hot object cools down to a temperature of 10 degrees of Celsius during a period of 180 seconds. The room temperature is 10 degrees of Celsius. Assume the coefficient in the Newton's Law of cooling to be 0.01[1/sec]. Determine the original temperature of the substance



    2. T(room) + (T(initial) T(room)) e^-kt = T(final)

    t room = 10
    ti = ?
    Tfinal = 10

    3. 10 + (Ti -10 ) e^(-0.01)(180) = 10
    10 + Ti - 10 + 0.1652 = 10
    0.1652 = 10 = 60.53

    I think this wrong because its a multiple choice and the choices are 93 and 100

    Please help me thank you!
     
  2. jcsd
  3. Dec 7, 2011 #2
  4. Dec 8, 2011 #3
    :( Sorry I still struggling with the problem, can you please show me how to do ? please!!
     
  5. Dec 8, 2011 #4
    You wrote,

    "10 + (Ti -10 ) e^(-0.01)(180) = 10"

    Subtract 10 from both sides and we have,

    (Ti -10 ) e^(-0.01)(180) = 0 --> Ti = 10


    The object did not cool down to room temperature in 180 seconds unless it started at room temperature. I think your facts are wrong. Assume that 100 was the starting temp and work backward to find the final temp, which can't be 10.


    Tf = 10 + (100 - 10)*exp(-1.8) Tf = 24.9

    Tf = 10 + (93 - 10)*exp(-1.8) Tf = 23.7

    Good luck!

    You might also visit,

    http://demonstrations.wolfram.com/NewtonsLawOfCooling/
     
    Last edited: Dec 8, 2011
  6. Dec 15, 2011 #5
    sorry, what is the ambient temperature? i've assignment where the question as follow:

    the temperature T of a cooling object drops at a rate proportional to the difference T-S, where S is constant temperature of surrounding medium. If initially T= 100 C, find the temperature of the cooling object at any time. anyone....please help me..
     
  7. Dec 16, 2011 #6
    fahanaam, refer to the rules of this forum.
     
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