Calculating Amount of Ice to Cool Tea to 65°C

[erok81]

Homework Statement

Your 200-g cup of tea is boiling-hot. About how much ice should your add to
bring it down to a comfortable sipping temperature of65°C . Assume that the ice is
initially at−15°C . The specific heat capacity of ice is 0.5cal g⋅°C , for water is 1 cal g⋅°C.
The latent heat for melting ice is 80cal g.

Homework Equations

Equation for latent heat: L=Q/m

The Attempt at a Solution

First I need to find the heat lost by the water. This is done using:

Q=c_wm_wΔT → (1 cal g⋅°C)(200 g)(65 C - 100 C)

Q=-7000 calories

Assuming no heat is lost to anything else during the process, Q_lost = Q_gained

So the ice cube gains the heat lost by the water, or 7000 calories.

Here is where I am stuck.

I tried using the latent heat equation directly (L=Q/m → m=Q/L) using the latent heat of melting ice a 80 cal/g, but this gave me the incorrect answer.

How does one figure out the mass? I tried subbing in the specific heat capacity for Q in the latent heat eqn, but then my masses cancelled. So that didn't work.

 

[haruspex]

What temperature will the ice end up at, and what are the stages the ice will go through to get there?

 

[erok81]

The ice ends up at the same final temp as the tea, so 65. And it will go from solid to a liquid. I tried throwing in the latent heat (H) for melting ice at 80 cal/g but it didn't work out correctly.

 

[haruspex]

The ice ends up at the same final temp as the tea, so 65. And it will go from solid to a liquid. I tried throwing in the latent heat (H) for melting ice at 80 cal/g but it didn't work out correctly.

How much heat per unit mass will be taken up by each of the three stages? (Pls show your working, not just 'it was wrong'.)

 

[erok81]

If I am understanding correctly (which I don't think I am)...

For melting ice 80 cal/g
For boiling water 540 cal/g

I don't have work for those as they are given in the text. Which is just the latent heat formula L=Q/m

I'm not sure on the three stages you mentioned. The ice will only go from frozen water to liquid water - so one phase transition.

All I tried above is using L=Q/m => m=Q/L => (7000 calories)/(80 calories/g) = 87.5g. But the correct answer is 40 something.

 

[haruspex]

The ice starts at what temperature? What temperature does it have to reach to even start melting? What temperature will it be at when it has just melted?

 

[erok81]

Ooooh. I see where you are headed now!

It starts at -15C. So I need to figure how much heat it's taking from -15 to zero, the latent heat part from freezing to non freezing, then 0C up to 65C it's final temperature.

And look at that; I just happen to have three stages just like you mentioned above. :)

So let's see where this takes us.

Still using m=Q/L

This is a guess, I'm not sure why it's like this. It just makes sense to put them all in the denominator since that is where all of the stages are happening.

[itex]m=\frac{Q}{(c_i * \Delta T) + L + (c_w * \Delta T)}[/itex]

Plugging in values...

[itex]m=\frac{7000}{(0.5 * 15) + 80 + (1 * 65)}[/itex]

Which equals 45.9g of ice.

Am I close to why all of the stages get thrown into the denominator?

L=Q/m to accomplish the transformation according to my text. So to accomplish what I am doing takes all three stages to make up L.

Assuming I've even done the solution correctly that is.

 

[haruspex]

That's it.

 

[erok81]

Thanks for the help. :)