Thermal Conductivity and Boundry Layer

  1. Thermal Conductivity , k , for water is .1455 cal/sec/ meter * C ( converted watts to cal/sec)
    Q/t = k* A delta T/ d
    A= area M2
    d = thickness of water/ice boundary layer
    Suppose I want to calculate time for 100 grams of ice to melt ( 8000 calories absorbed
    from surroundings by conduction.
    I have delta T and initial surface area of ice/water interface .
    Two questions : What would be the boundary layer thickness. d
    And in the case that the surface area, A, is decreasing as the ice melts is it correct to
    use just the initial surface area in the calculation ?
    My reasoning on the area is that it would actually be : Integral A(initial) - A(final) and
    A(final ) = 0
     
    Last edited: Apr 10, 2011
  2. jcsd
  3. I just did this experiment with 100 cc ice in 1 liter water at 30 deg C , 30000 cal
    so final temp 22000 cal /1100 gram water = 20 C , melting time was 6 minutes
    In close agreement with Q/t = k*A delta T/ x
    x = thickness of boundary layer between ice and water = .002 M
    Reference Journal Physical Oceanography
    A= surface area 100 cc ice cylinder = .01224 M^2
    delta t = 30 deg
    k, thermal conductivity water = .1455 cal/sec/Meter * Kelvin
    So, Q/t = .1455 cal/sec/M*T ( 30 C) (.01224 M^2) / .002 M = 26.5 cal/sec
    8000 cal/ 26.5cal/sec = 301 sec. = 5 min
     
    Last edited: Nov 24, 2011
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