Thermodynamics: phase change and latent heat problem

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Homework Statement:
What mass of steam initially at 130°C is needed to warm 200 g of water in a 100-g glass container from 27.0°C to 42.0°C? If we instead used 78.0 g of water, how much 100°C steam remains in thermal equilibrium?
Relevant Equations:
Q = mc(delta)T, Q = L(delta)m, Q(cold) = -Q(hot)
c(water)= 4186 J/kg*C, c(steam)= 2010, c(glass)= 837. L(water, vaporizing)= 2.26e6 J/kg
The second question is where I'm lost. The answer to the first question is 5.39 grams. The second is 10.3 grams. Until I saw the answer I was setting the equation up as if some of the original steam had condensed. But it appears that some of the original water changed into steam. Is it necessary to know before hand whether its the water or steam that undergoes the phase change? My attempt at a solution was to set Q(cold)=-Q(hot).

subscripts.... w = water, s = steam, g = glass, v = vaporization

ch20_prob4.jpg
 

Answers and Replies

  • #3
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Thanks for the reply. Im not sure I understant how latent heat works in this scenario. In a system like this, with water and steam, if one undergoes a phase change, does the latent heat affect the temperature of the other? For example, the water stays at 100C until it has completely changed to steam, but does the initial steam stay at whatever temp it was when the phase change started? Hope that makes sense...
 
  • #4
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If the liquid water is changing to steam at 100 C, it is receiving heat to do this. The best thing to do in a problem like this is to think of the initial water and the initial steam as separate entities, and calculating how much heat must be added or received from each of them separately to arrive at the final equilibrium state.
 
  • #5
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OK, I think I get it. The water hits 100C and starts to undergo phase change. As soon as the initial steam hits 100C the system is in equilibrium. I'll see if i can work that out. Thanks for your help!
 
  • #6
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I think I figured it out. Or figured out a way to arrive at an answer within 1/10th of a gram lol. Whereas the first part of the problem required you to include the heat necessary to raise the temperature of the glass, I could only arrive at the "correct answer" for the second part by not including the glass.

As you can see, theres quite a deficit on the right side of the equation. So Initial steam has to undergo phase change to raise the temperature of the water to 100C. Thanks for all your help Chestermiller!

IMG_2683[607].jpg
 
  • #7
haruspex
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The 2nd problem makes no sense to me.
I think the second part intends we use the same mass of steam as calculated in the first part, and implies that all will reach 100C before all the steam condenses. Unfortunately, that is not the case.
I think I figured it out. Or figured out a way to arrive at an answer within 1/10th of a gram lol. Whereas the first part of the problem required you to include the heat necessary to raise the temperature of the glass, I could only arrive at the "correct answer" for the second part by not including the glass.

As you can see, theres quite a deficit on the right side of the equation. So Initial steam has to undergo phase change to raise the temperature of the water to 100C. Thanks for all your help Chestermiller!
Your calculation has found the mass of steam that needs to condense. That is not what was asked. To answer the question as given, you have to know how much steam you started with, and the only figure we have for that is 5.39g - less than the mass needed to raise all the water to 100C, and even then ignoring the need to warm the glass.

I wondered whether the second part was supposed to say "if we had enough steam to raise 200g of water (and the glass) from 27C to 100C, but there is only 78g then how much steam will remain?", but I get 16.5g for that.
 
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  • #8
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Maybe the problem statement meant to say that 78 grams of steam are added, rather than water.
 
  • #9
haruspex
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Maybe the problem statement meant to say that 78 grams of steam are added, rather than water.
But then the mass left as steam would be about 73g, not the given 10.3.
 
  • #10
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I think the second part intends we use the same mass of steam as calculated in the first part, and implies that all will reach 100C before all the steam condenses. Unfortunately, that is not the case.

Your calculation has found the mass of steam that needs to condense. That is not what was asked. To answer the question as given, you have to know how much steam you started with, and the only figure we have for that is 5.39g - less than the mass needed to raise all the water to 100C, and even then ignoring the need to warm the glass.

I wondered whether the second part was supposed to say "if we had enough steam to raise 200g of water (and the glass) from 27C to 100C, but there is only 78g then how much steam will remain?", but I get 16.5g for that.
Your right, my answer makes no sense with regard to initial amount of steam. I think Id had my head buried in the numbers for too long at that point.
 
  • #11
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I think I've arrived at the conclusion that its just a poor problem. The homework site im using will generate a similar problem for practice, with slight changes to initial variables. Heres another- solving the second part without the glass results in a logically consistent solution.

ch20_4.png
ch20_4_1.png
ch20_4_2.png
ch20_4_solution.jpg


Theres a quote in my textbook- "Δm always refers to the higher phase material." So im operating under the assumption that Δm is the amount of steam remaining. Though this doesnt really make sense to me in the context of mf - mi = Δm.
 
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  • #13
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That's not what I get. I get 50.3 g steam remaining.
Ok, but still no single-error explanation for the textbook answer.
Δm always refers to the higher phase material." So im operating under the assumption that Δm is the amount of steam remaining.
I would interpret it as saying Δm is the change in the higher phase material. Unclear whether that is as a magnitude or a scalar.
 

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