Heat transfer calorimeter question

[borusik]

Homework Statement

A copper calorimeter with mass 0.1 kg contains 0.15 kg of water and 0.012 kg of ice in their equilibrium at normal atmospheric pressure(0 degrees Celsius). If 0.5 kg of lead at 200 Celsius is dropped in calorimetr, what is the final temperature?( consider the close system so no temperature lost).

Homework Equations

Cw*Mw*(Tf-Ti)=Cice*Mice*(Tf-Ti)=Ccop*Mcop*(Tf-Ti)=Clead*Mlead*(Ti-Tf)

The Attempt at a Solution

I'm not sure if i have to consider the latent phase, so any help would be appreciated.
Thank you.

 

[dynamicsolo]

There is some ice in the initial state of the calorimeter, so you will need to include a term for the latent heat of melting the 12 g. of ice.

Oh, and your equation is not going to be of any help to you like that. All the terms can't be equal! A standard way to set up a calorimetry equation is to put all the items that will be releasing heat on one side and the items which will be receiving heat on the other, with Qout = -Qin .

 

[borusik]

Thank you dynamicsolo. Is it possible for you to tell me the number of ice latent heat of fusion?

 

[dynamicsolo]

For you, certainly! It's 334 J/gm = 79.8 cal/gm for ice.

But keep in mind that Google is your friend (as they say on many forums). "Everything" is on the 'Net these days; for instance:

http://en.wikipedia.org/wiki/Latent_heat (see the table under 'Water')