Calorimetry: Finding mass of ice, water, steam left

In summary, the problem involves finding the final temperature of a copper calorimeter containing 95 grams of ice at 0.0°C and 35 grams of steam at 100.0°C and 1.00 atm pressure. Using the equation Qloss = Qgain, the final temperature is calculated to be 86.47°C. Since this temperature is above the melting point of ice, all of the ice is melted. The remaining mass of steam is found to be 21.8 grams, while the mass of liquid water left is 108.2 grams. The calculation for Tf assumes that all steam condenses.
  • #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):
Qloss = Qgain
Where Steam = Calorimetry + Ice
msLv + mscs(100°C) - mscsTf = mcccTf - mccc(0°C) + miLf + miciTf - mici(0°C)
Arrange the equation to find Tf
Tf = miLf - msLv - mscs(100°C) / (- mscs - mccc - mici)
Tf = (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°))
Tf = 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?
Qice = (95g)(80cal/g) = 7600 cal
Qsteam = (35g)(540cal/g)-(35g)(1cal/g⋅C°)(86.47°C)+(35g)(1cal/g⋅C°)(100.0°C) = 19373.55 cal
Since Qsteam > Qice, all ice is melted.

For mass of steam left:
7600 cal = ms(540cal/g)-(35g)(1cal/g⋅C°)(86.47°C)+(35g)(1cal/g⋅C°)(100.0°C)
ms = 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
 
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  • #2
paulie said:
Since the temperature is around 86.47°C, doesn't it mean that the ice are all melted?
Yes.
paulie said:
For mass of steam left:
Didn't your calculation of Tf assume all steam condenses?
 

Related to Calorimetry: Finding mass of ice, water, steam left

1. How is the mass of ice, water, and steam determined using calorimetry?

Calorimetry is a scientific method used to determine the heat transfer that occurs during a physical or chemical process. In this case, the mass of ice, water, and steam can be determined by measuring the amount of heat absorbed or released during a phase change.

2. What equipment is needed for calorimetry?

The basic equipment needed for calorimetry includes a calorimeter, a thermometer, and a balance. A calorimeter is a device used to measure the heat transfer during a process, while a thermometer is used to measure the temperature change. A balance is used to measure the mass of the substances being studied.

3. How is the experiment set up for calorimetry?

To set up the experiment for calorimetry, the calorimeter is filled with a known amount of water and the initial temperature is recorded. Then, the substance being studied (ice, water, or steam) is added to the calorimeter and the change in temperature is measured. The change in temperature and the known heat capacity of water can be used to calculate the amount of heat absorbed or released during the process.

4. What is the purpose of determining the mass of ice, water, and steam using calorimetry?

The purpose of determining the mass of ice, water, and steam using calorimetry is to understand the heat transfer and phase changes that occur during a physical process. This information is important in many areas of science, including thermodynamics, chemistry, and engineering.

5. Are there any limitations to using calorimetry to determine the mass of ice, water, and steam?

While calorimetry is a useful tool for determining the mass of ice, water, and steam, there are some limitations to consider. For example, the accuracy of the measurements can be affected by external factors such as heat loss to the surroundings or incomplete phase changes. Additionally, the heat capacity of the substances being studied may vary slightly, which can also affect the accuracy of the results.

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