
#1
Oct212, 07:54 AM

P: 49

1. The problem statement, all variables and given/known data
A small pond has a layer of ice 1cm thick floating on its surface. The air temperature is 10 C. Steps (ac) should help you fi nd the final answers in (d) and (e) (a) What is the temperature of the liquid water just below the ice? (b) Write the expression for the rate of heat transfer through the ice as a function of thickness of the ice? (c) Write the expression for the rate of heat flow required to freeze the water? (d) Find the initial rate in cm/hr at which ice is added to the layer. (e) How long does it take for a 20cm layer to build up? . 2. Relevant equations H= kA (ThTc)/L H=dQ/dT 3. The attempt at a solution For A) im not sure of the formula to use, I know only H = kA (ThTc)/L, but the question only provides L and Th and expects me to find Tc somehow. For B) and C) i'm not really sure what it means by an expression (like dQ/dt = kA (ThTc)/L?) would B) be like dH(L) = kA(ThTc)? For D) i have the save problem as in question A) where I do not feel I have enough variables to solve for dQ/dt i think i can probably solve for e) once I get the other parts. Please offer any help you can! thanks! 



#2
Oct212, 08:28 AM

P: 1,195

"For A) im not sure of the formula to use, I know only H = kA (ThTc)/L, but the question only provides L and Th and expects me to find Tc somehow."
The ice forms at the waterice interface. There is a latent heat that must be removed from the water once it reaches 0 C. So what does that say about the temperature of the water just below the ice? "For B) and C) i'm not really sure what it means by an expression (like dQ/dt = kA (ThTc)/L?) would B) be like dH(L) = kA(ThTc)?" Plug in the units and you'll see what it describes. There are several modes of heat transfer. One is conduction, one is convection, and the third is radiation. Conduction is heat diffusing through a material, generally a solid. Convection is heat transfer between a fluid (gas or liquid) and a solid. Radiation is energy transport by waves which is not considered in this problem. Any heat that leaves the surface of the ice and passes to the air is conducted through the ice. The heat that is conducted through the ice is the latent heat of solidification. 



#3
Oct212, 09:43 AM

P: 49

also for b) does the expression of rate of heat flow through the ice as a function of thickness just mean it starts with dH(L)=.....? I'm really unfamiliar with writing expressions so I have no idea. 



#4
Oct212, 10:35 AM

P: 1,195

heat transfer (conduction questions)
"Does that mean for a) the water below the ice would be just above 0 degrees? Do you not need a formula or is there a derived version of the one I have that I need to use?"
The water that is still liquid that is in contact with the ice is at 0 C. Water only freezes at 0 C at atmospheric pressure. It remains at that temperature until it completes the freezing process by removal of latent heat. "also for b) does the expression of rate of heat flow through the ice as a function of thickness just mean it starts with dH(L)=.....? I'm really unfamiliar with writing expressions so I have no idea." Conduction of heat through a solid is expressed as: Q = k*A(T1  T2)/L where the temperatures are the temperatures a distance L apart. k is conductivity. A is the area. The units of Q are energy per unit time. Any heat that is conducted through the ice is lost to the atmosphere via convection. Its formula is: Q = h*A*(TiceTair). h is the film coefficient. Q has the same units as above. As the ice builds up, it offers more resistance to heat loss. Thus the rate of ice formation is not constant in time. It occurs faster when the ice is thin. You have two resistances to heat flow here. One is conduction, the other is convection. When taken together they can be written as: Q/A = (T1  T2)/(1/h + L/k) which represents the heat transfer per unit time and area and it equates to the rate of ice formation. The difficulty in this problem is that L is a function of time because it is the ice thickness. Another difficulty of this problem is that when the ice gets thick, some of it may be below the freezing point. What level course is this? 



#5
Oct212, 11:33 AM

P: 49

This is only first year phyiscs




#6
Oct212, 11:42 AM

P: 49





#7
Oct212, 11:55 AM

P: 1,195

Can you solve an ordinary differential equation? If so here is how it would go.
The rate at which heat needs to pass to the atmosphere is G * A * dL/dt where G is the latent heat of freezing per unit volume. A is area; dL/dt is time rate of ice formation. If you equate this to the total resistance of the ice and film coefficient resistances times overall temperature difference you get: G * A *dL/dt = A *(Tice  Tatm)/(1/h + L/k) Rearranging and cancelling out A: G *dL/dt = k * h * (Tice  Tatm)/(k + h*L) Tice = 0; Tatm=10 The above is an ordinary differential equation for the ice thickness. Solve it and you will get the ice thickness as a function of time. L is the dependent variable; time is the independent variable. 



#8
Oct212, 12:05 PM

P: 1,195

"Is it possible to solve part b) while neglecting convection? The reason I ask is because while we have gone over convection there is no mention of any formulas or calculations involving convection in the textbook or any of the lectures. I think our professor specifically noted that we wouldn't bother with convection other than knowing what it is so I don't feel that homework assigned would cover it."
OK, so no convection. Then assume the ice surface is at the atmospheric temperature of 10 C. Mathematically, that means that h is infinite. G *dL/dt = (Tice  Tatm)/(1/h + L/k) for infinite h, the equation reduces and is easier to solve because the 1/h terms drops out. In passing, whenever there is a huge film coefficient, it essentially forces the surface temperature of the object to its value. Hence, we assign the value of the ice surface to 10 C. 



#9
Oct212, 12:08 PM

P: 49

I'm not quite sure where you get the equation
H=G*A*dL/dt Also are heat flow and heat transfer the same thing? 



#10
Oct212, 12:33 PM

P: 1,195

As for the term GA dL/dt, it represents a rate of heat transfer to freeze water. If you had a volume of liquid water at 0 C, the energy you'd have to remove from it to turn it all to ice (also at 0 C) would be G times its volume where G is the heat of fusion needed per unit volume. So the rate at which this is done is G * A * dL/dt where A is area, and dL/dt is the rate of thickness change of the layer. Note the units. They work out to be Joules/hour. They agree with the right hand side of the equation. A*dL/dt is rate of volume change. 


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