Question 1066970
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A tank of water has been outdoors in cold weather until a 6.00 cm thick slab of ice has formed on its surface. 
The air above the ice is at -17.0 degrees C. Calculate the rate of formation of ice (in centimeters/hour) on the bottom surface
 of the ice slab. Take the thermal conductivity of ice to be 0.0040 cal/s-cm-degree C, the density to be 0.92 g/cm^3, 
and the heat of fusion to be 80 cal/g. Assume that no heat enters or leaves the water through the walls of the tank.
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The units are not SI, but they all are consistent (!), so we can make all calculations from the beginning to the end 
without making conversion the units (thanks to God).


<pre>
1.  The general balance equation is 

    {{{rho*phi*(dL/dt)}}} = {{{lambda*((DELTA(T))/L)}}}.


    The right side is the heat flow by conductivity from the water in the tank to the atmosphere through the ice layer: 
    
    DELTA(T) is the temperature difference of 17 °C across the ice layer;  
             (we assume that the water temperature in the tank is 0 °C under the ice layer)

    L = 6 cm is the layer thickness;

    {{{lambda}}} = 0.004 {{{cal/(s*cm_*degC)}}} is the thermal conductivity of ice.



    The left side is the amount of heat to increase the thickness of the ice layer:

    {{{(dL)/(dt)}}} is the rate of the thickness increasing, in {{{cm/s}}}; it is the value and the quantity under the problem question;

    {{{phi}}} is the specific latent heat of freezing, 80 {{{cal/g}}};

    {{{rho}}} is the density of ice, 0.92 {{{g/cm^3}}}.


    You may check that the dimensions of both sides are consistent.
    If you never did/check it before, make this very useful exercise. 


2.  Now substitute all given data to get

    
    {{{(dL)/(dt)}}} = {{{(0.004*(17/6))/(0.92*80)}}} = 0.000154 {{{cm/s}}} = 0.554 {{{cm/hour}}}.
</pre>

<U>Answer</U>.  The rate of formation of ice is  0.554 {{{cm/hour}}} under the given condition.