Calorimetry: Finding mass of ice, water, steam left

[paulie]

Homework Statement

Given a copper calorimeter that has a mass of 446 grams containing 95 grams of ice at 0.0°C. If 35 grams of steam at 100.0°C and 1.00 atm pressure is added to the can, what is the final temperature of the can and its contents? At the final temperature, how many kilograms are there of ice, how many of liquid water, and how many of steam?

Homework Equations

Q=mcΔT

The Attempt at a Solution

Finding the final temperature (I think I got this part right already but here it is):
Q_loss = Q_gain
Where Steam = Calorimetry + Ice
m_sL_v + m_sc_s(100°C) - m_sc_sT_f = m_cc_cT_f - m_cc_c(0°C) + m_iL_f + m_ic_iT_f - m_ic_i(0°C)
Arrange the equation to find T_f
T_f = m_iL_f - m_sL_v - m_sc_s(100°C) / (- m_sc_s - m_cc_c - m_ic_i)
T_f = (95g)(80cal/g)-(35g)(540cal/g)-(35g)(1cal/g⋅C°)(100.0°C) / (-(35g)(1cal/g⋅C°)-(446g)(0.0923cal/g⋅C°)-(95g)(1cal/g⋅C°))
T_f = 86.47°C

How do I find the mass of ice, water, and steam left?
Since the temperature is around 86.47°C, doesn't it mean that the ice are all melted?
Q_ice = (95g)(80cal/g) = 7600 cal
Q_steam = (35g)(540cal/g)-(35g)(1cal/g⋅C°)(86.47°C)+(35g)(1cal/g⋅C°)(100.0°C) = 19373.55 cal
Since Q_steam > Q_ice, all ice is melted.

For mass of steam left:
7600 cal = m_s(540cal/g)-(35g)(1cal/g⋅C°)(86.47°C)+(35g)(1cal/g⋅C°)(100.0°C)
m_s = 13.20 grams (amount of steam condensed)
35g - 13.20g = 21.8 grams (amount of steam left)

For the mass of liquid left:
13.20 grams (amount of steam condensed) + 95grams (melted ice) = 108.2 grams

 

[haruspex]

Since the temperature is around 86.47°C, doesn't it mean that the ice are all melted?

Yes.

For mass of steam left:

Didn't your calculation of T_f assume all steam condenses?