Clarifying Latent Heat: Cooling 1kg of Water to 0C?

[PrudensOptimus]

There was this problem regarding Latent Heat that I would like further clarification:

Problem:

200g of ice at -10C is added to 1kg of water at 15C in an insulated container. Is there enough ice to cool the water to 0C? If so, how much ice and water are present once equilibrium is reached?

In the book it demonstrated the solution by finding the Energy required to melt all 200g of ice Q1:

Q1 = mc(dT) + mL = 70.9kJ -- I understand that.

Then it says "Cooling the water to 0C extracts an amount of heat given by Q2":

Q2 = mc(dT) = 1kg(4.184kJ/kg*K)(15K) = 63kJ. -- Why did they set initial Temperature to be 0C?

 

[Hurkyl]

They're calculating a negative (the heat extracted), so the equation has an extra negative sign.

In particular, $- \Delta T = -(T_f - T_0) = T_0 - T_f$

 

[PrudensOptimus]

OK, when they calculate the mass of the water at 0C after apply ice, they used this:


m = Q/L = 58.7kJ/334 = 176g.

I understand they subtracted 4.1kJ(Specific heat of Water) from 62.8kJ... But why did they do that?

 

[selfAdjoint]

Because in order to freeze, to pass from the liquid state to the solid state without any change in temperature, the sample had to LOSE that much heat to the environment (by conduction or radiation).

Think of it as representing the amount of kinetic energy lost by the molecules between being able to move around, as in the liquid, and being fixed in a crystilline lattice, in the solid.

The heat has to go somewhere, and it goes into the environment. That is why you get the paradoxical result that when, say, droplets in the air freeze to ice crystals, they have a warming effect on their part of the atmosphere. Conversely when ice crystals melt they take up that specific heat from the environment, resulting in a cooling effect on the surrounding air.

Remember that these exchanges of specific heat don't in themselves change the temperature of the water; this heat goes only into the change of state.

 

[PrudensOptimus]

Is there another explanation for Q2?

And perhaps someone can work this problem in a slight different, clear, simple way?

 

[Doc Al]

Originally posted by PrudensOptimus
I understand they subtracted 4.1kJ(Specific heat of Water) from 62.8kJ... But why did they do that?

Think of the process in steps. Before you can melt the ice, you must first warm it up to the melting temperature. You know that the energy needed to:
1) warm the ice to 0 = 4.2kJ
2) melt all the ice (at 0) = 66.8kJ

The available energy in the warm water (compared to 0 degrees) is: 63kJ

That's more than enough to warm the ice, but not enough to melt it all. (So we know the final temperature will be 0.) After warming the ice, there will be 63-4.2=58.8kJ left to melt ice. So a fraction equal to 58.8/66.8 (= 0.88) of the ice is melted (= 176g).

 

[PrudensOptimus]

Originally posted by Doc Al
Think of the process in steps. Before you can melt the ice, you must first warm it up to the melting temperature. You know that the energy needed to:
1) warm the ice to 0 = 4.2kJ
2) melt all the ice (at 0) = 66.8kJ

The available energy in the warm water (compared to 0 degrees) is: 63kJ

That's more than enough to warm the ice, but not enough to melt it all. (So we know the final temperature will be 0.) After warming the ice, there will be 63-4.2=58.8kJ left to melt ice. So a fraction equal to 58.8/66.8 (= 0.88) of the ice is melted (= 176g).

I see, all previous responses in combine with your response there made me understood the problem completely. I am very greatful and I will cogitate on the procedures now. Thanks.