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Homework Help: Time to take for water to cool down.

  1. Apr 13, 2012 #1
    My question refers to the cooling of water in a well-insulated , cylindrical cup that is opened at the top.

    0.336Kg of water at 84degCel is poured into the cup which is left to stand on a table at an ambient temperature of 25degCel. The specific heat capacity of water is 4150 J Kg^-1 K^-1 and the U-value is 10.64Wm^-2K^-1. The area of the water surface that is exposed to the surroundings air is 0.00447 m^2.

    So, how to calculate the time taken in seconds for the water to cool down to 46degCel?

    This is what I have done so far....

    Let x = (U-value)(Surface Area)/(mass)(Heat Capacity)
    Let y1 = 46DegCel (cool down temp)
    Let y2 = 84DegCel (Water temp)
    Let y3 = 25DegCel (Ambient tempt)
    Taking time to be T, so...

    T = -1/x log (y1-y3)/(y2-y3)
    = -13153 (sec)

    I got negative, so my guess something is wrong here. The formula is from my book. Need help....
  2. jcsd
  3. Apr 14, 2012 #2
    Have you studied Newtons Law of Cooling?
  4. Apr 14, 2012 #3
    I saw the mistake...it's suppose to be log base e thus the negative is cancelled out. Btw, ya I've studied Newton's Law
  5. Apr 14, 2012 #4


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    Staff: Mentor

    Couple of things. First, the log should be the natural logarithm, loge not log10. Second, your argument for the log is less than one, so it should yield a negative value that is subsequently dealt with bu the "-" in front of 1/x.
  6. Apr 14, 2012 #5
    sorry i wasnt able see newtons law of cooling equation. You have written it in such a way that i thought you have written some other thing. I saw it now
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