Mixing water with steam, find equilibrim

AI Thread Summary
When mixing 100 g of steam at 100 °C with 500 g of water at 25 °C, the steam will condense as it transfers heat to the cooler water, reaching thermal equilibrium. The heat lost by the steam during condensation is 226,000 J, while the maximum heat the water can absorb is 156,975 J. Since the water cannot absorb all the latent heat, some steam will condense, increasing the total water mass to 600 g. The process is known as direct contact condensation, where steam condenses until the temperatures equalize. Ultimately, the final temperature will stabilize at 100 °C when equilibrium is reached.
sallychan
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Homework Statement


100 g of steam at 100 °C is mixed with 500 g of water at 25 °C. What is the final temperature of the mixture?

Homework Equations



Q= mL

Q= mcΔT

The Attempt at a Solution



Heat given by to water = heat loss at condensation
= mLv
= (0.1) (2.26 E 6)
= 226000 J

Maximum possible heat water can take = mc ΔT
= (0.5) (4186)(100-25)
= 156975 J

Because water does not take up all the latent heat from steam, only some of the steam condensate.

I wonder is my conclusion right? Or will latent heat given by steam actually boils water?

Thanks!
 
Last edited:
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Your conclusions are correct. Congratulations for having the intuition that 100degC is the maximum temperature possible.
 
But how am I going to find the final temperature if all water is boiled?
 
Think about a 500g pot of water with a tube going to the bottom. Through that tube you inject the 100g steam. What will you see after the water reaches 100C?
 
insightful said:
Think about a 500g pot of water with a tube going to the bottom. Through that tube you inject the 100g steam. What will you see after the water reaches 100C?
Theoretically, when water reaches 100C, they become steam?
 
Picture a bubble of 100C steam entering below the surface of 100C water. What happens to that bubble?
 
insightful said:
Picture a bubble of 100C steam entering below the surface of 100C water. What happens to that bubble?
I think the bubbles will float on the surface of the water... just guessing
 
Good guess! So the bubbles come off the surface carrying the same amount of water that came into the pot with the steam. How much of the water (liquid) phase leaves the pot?
 
insightful said:
Good guess! So the bubbles come off the surface carrying the same amount of water that came into the pot with the steam. How much of the water (liquid) phase leaves the pot?
Oh so you are saying that the steam will condensate? If that is the case, the total amount of water will become 500g + 100g = 600 g.
But how does that happen? Isn't it water does not remove enough latent heat from the steam?
 
  • #10
Well, no. For the steam to condense, the water must be colder than the steam. Think about it for a while.
 
  • #11
insightful said:
Well, no. For the steam to condense, the water must be colder than the steam. Think about it for a while.
Oh right. I understand.
But can you explain a bit more on the heat exchange of condensation? I really want to fully understand it.
Thanks a lot!
 
  • #12
Your problem is called direct contact condensation. As you have realized, the hot steam will condense into the cold water until there is no temperature difference. You have reached equilibrium between the steam and the water at that point. That's really all there is to it.
 
  • #13
insightful said:
Your problem is called direct contact condensation. As you have realized, the hot steam will condense into the cold water until there is no temperature difference. You have reached equilibrium between the steam and the water at that point. That's really all there is to it.
So the steam will always condense when in contact with water in lower temperature?
 
  • #14
You got it.
 
  • #15
insightful said:
You got it.
Okay Thank you very much!
 
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