Final temperature of system containing ice and a copper can when steam is added?

[JustinLiang]

Homework Statement

A copper calorimeter can with mass 0.466kg contains 00950kg of ice. The system is initially at 0.0C. If 0.0350kg of steam at 100C and 1atm pressure is added to the can, what is the final temperature of the calorimeter can and its contents?

Homework Equations

Q=mL
Q=mcΔT

The Attempt at a Solution

I am going to assume all the ice melts and all the steam boils. If my answer is between 0C and 100C then I know that this is true. If it is below 0C or above 100C I will have to solve it differently.

Thus we have
-->Q(copper) + Q(ice melt) + Q(water warming) = Q(steam condensing) + Q(water cooling)
-->(0.446kg)(390)(Tf-0) + (0.095kg)(334000) + (0.095kg)(4190)(Tf-0) = (0.035)(2256000) + (0.035)(4190)(Tf-100)
When I solve this I get 76.6C, however the answer is 86.1C.

What am I doing wrong?

 

[ehild]

-->Q(copper) + Q(ice melt) + Q(water warming) = Q(steam condensing) + Q(water cooling)
-->(0.446kg)(390)(Tf-0) + (0.095kg)(334000) + (0.095kg)(4190)(Tf-0) = (0.035)(2256000) + (0.035)(4190)(Tf-100)


What am I doing wrong?

The red one. It should be 100-T, as the term means the heat added to the ice, a positive quantity.

ehild

 

[JustinLiang]

The red one. It should be 100-T, as the term means the heat added to the ice, a positive quantity.

ehild

I always see it as change in temperature? When you put it as 100-T doesn't that mean the final temperature is 100 and the initial is Tf? I am still very confused :S

 

[ehild]

You wrote all terms required to warm up the ice and calorimeter on the left hand side, and the heat amounts available on the right-hand side. These are the heat released at deposition and the heat released during the cooling of the hot water, proportional to 100-Tf, otherwise it would be negative.

ehild