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

In summary, the conversation discusses a problem involving a copper calorimeter can with ice, and the addition of steam at 100C and 1atm pressure. The problem asks for the final temperature of the can and its contents. The equations used to solve the problem are Q=mL and Q=mcΔT, and the solution involves setting up an equation with all the heat terms on the left side and the heat amounts available on the right side. The final temperature is found to be 76.6C, but the correct answer is 86.1C. The error is identified as using 100-T instead of 100+T in one of the terms.
  • #1
JustinLiang
81
0

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?
 
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  • #2
JustinLiang said:
-->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
 
  • #3
ehild said:
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
 
  • #4
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
 
  • #5

As a scientist, it is important to carefully consider all the variables and assumptions in a problem before attempting to solve it. In this case, there are a few things that need to be clarified.

Firstly, it is important to note that the specific heat capacity of copper (390 J/kgK) and water (4190 J/kgK) are different, so they cannot be treated as the same variable in your equation. Additionally, the latent heat of fusion (334000 J/kg) and latent heat of vaporization (2256000 J/kg) should also be taken into account.

Secondly, you have correctly identified that the final temperature could be between 0C and 100C, depending on whether all the ice melts and all the steam boils. However, it is important to consider what happens if the final temperature is outside of this range.

If the final temperature is below 0C, it means that some of the water will still be in the form of ice. In this case, you would need to include the energy required to melt the remaining ice in your equation.

If the final temperature is above 100C, it means that some of the water will still be in the form of steam. In this case, you would need to include the energy required to vaporize the remaining water in your equation.

Taking all of these factors into consideration, a more accurate approach would be to set up separate equations for the different scenarios (i.e. final temperature between 0C and 100C, below 0C, and above 100C) and solve for the final temperature in each case. Then, compare the results to determine which scenario is the most likely outcome.

In conclusion, it is important to carefully consider all variables and possible outcomes when solving a scientific problem. In this case, a more thorough analysis and consideration of all the relevant factors may lead to a more accurate and satisfactory answer.
 

1. What is the final temperature of the system when steam is added?

The final temperature of the system will depend on the initial conditions, such as the amount of ice and the temperature of the copper can. However, the final temperature will be between 0 degrees Celsius (the melting point of ice) and 100 degrees Celsius (the boiling point of water).

2. How does the addition of steam affect the final temperature of the system?

The addition of steam will increase the temperature of the system. This is because the steam has a higher temperature than the ice and will transfer heat to the system.

3. What factors influence the final temperature of the system?

The final temperature of the system is influenced by the initial conditions, such as the amount of ice and the temperature of the copper can. It is also affected by the amount of steam added and the rate of heat transfer between the steam and the system.

4. Will the final temperature of the system be the same every time?

No, the final temperature of the system will vary depending on the initial conditions and the amount of steam added. In a controlled experiment, the final temperature can be predicted, but in real-world situations, it may vary.

5. Can the final temperature of the system exceed 100 degrees Celsius?

No, the final temperature of the system cannot exceed 100 degrees Celsius as this is the boiling point of water. If the system reaches this temperature, the steam will continue to transfer heat until it fully condenses into liquid form.

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