- #1

paulie

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## 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

_{s}L

_{v}+ m

_{s}c

_{s}(100°C) - m

_{s}c

_{s}T

_{f}= m

_{c}c

_{c}T

_{f}- m

_{c}c

_{c}(0°C) + m

_{i}L

_{f}+ m

_{i}c

_{i}T

_{f}- m

_{i}c

_{i}(0°C)

*Arrange the equation to find T*

_{f}T

_{f}= m

_{i}L

_{f}- m

_{s}L

_{v}- m

_{s}c

_{s}(100°C) / (- m

_{s}c

_{s}- m

_{c}c

_{c}- m

_{i}c

_{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