How Much Heat is Needed to Turn Ice at -10°C into Steam at 110°C?

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Homework Help Overview

The problem involves calculating the total heat required to convert 150g of ice at -10°C into steam at 110°C, incorporating various phase changes and temperature increases. The subject area includes thermodynamics and heat transfer principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the necessary equations and processes involved, including the specific heat capacities and latent heats for ice, water, and steam. Some express uncertainty about how to progress from initial calculations to the final answer, while others suggest considering the multiple stages of heating and phase changes.

Discussion Status

Several participants have provided calculations for specific parts of the problem, such as the melting of ice and heating of water. There is acknowledgment of the complexity of the problem, with some guidance on the need to sum the contributions from each stage of the process. However, there is no explicit consensus on the final approach or solution.

Contextual Notes

Participants note the importance of understanding the phase changes and the energy required for each step, as well as the potential for confusion regarding the calculations involved. There is a reminder to keep discussions focused on the original problem without diverging into unrelated topics.

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Homework Statement



Calculate the amount of heat to be added to warm 150g of ice from -10C to produce steam at 110C.

Homework Equations



Ice = cice = 0.5cal/gC, L = 80 cal/g.
Water = Cw = 1.0 cal/gC;
Steam = csteam = 0.5 cal/gC; Lv = 540 cal/g

The Attempt at a Solution



I know the answer should be 109.5kcal.

I have found 12,000 cal. BUT I HAVE ABSOLUTELY NO IDEA HOW TO GET TO 109kcal. I JUST NEED SOME DIRECTION.
 
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Show how you got 12 kcal.
 


Q = mLf = 150g x 80 Cal/g = 12,000
 


I know this problem is more complex than that.

Another equation that is possible to use is q=mcdeltat

I remember my professor drawing a graph, with equations for solid, liquid and gas...
I think I am supposed to have that in mind.
 


The graph shows increase, then constant, then increase and constant, then increase.
 


domyy said:
Q = mLf = 150g x 80 Cal/g = 12,000

That's just a melting part.

domyy said:
Another equation that is possible to use is q=mcdeltat

I remember my professor drawing a graph, with equations for solid, liquid and gas...
I think I am supposed to have that in mind.

domyy said:
The graph shows increase, then constant, then increase and constant, then increase.

And that's a good hint at equations that should be used. Have you copied that graph?

What must happen to the ice so that water becomes a steam?
 


There are 5 processes at work here. First, you must heat the water from -10C to 0C. Then more energy must be provided to turn the 0C ice to 0C water. Then you must heat the water from 0C to 100C. Then you must turn the 100C water to 100C steam. Finally, you heat the steam to the final 110C. Phase changes require energy as well as temperature changes. The equations you have can be used to determine exactly how these processes occur.
 


One of the solutions would be q=mlf
=150g x 540=81,000?
 


I will also have q=150g x 80cal/g = 12,000
 
  • #10


Am i doing it right? I will have to sum the values later?
 
  • #11


You can sum them all as the last step.
 
  • #12


Oh yes I already got the answer. Thanks a lot!
 
  • #13


Can I ask you something? I read that as bubbles rise in deep column of water the diameter increases. Could explain this more clear to me?
 
  • #14


Please don't hijack threads, if you have a new question - start a new thread.

PV=nRT, P=ρh
 

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