Entropy of water-ice bath

  • Thread starter Metaleer
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Hey, all.

I'm tutoring a freshman, and we've come to this particular problem, and I think there's missing data. He does have access to the vapor tables, but as you can see, he's not given total mass, volume, nor any specific heat... here goes:

"An adiabatic calorimeter contains a water-ice mixture at 0°C and 1 atm. Vapor at 100° is introduced in the calorimeter until the mixture has gained 1.80 g of liquid water, the final temperature of the mixture being 0°C. If the latent heat of evaporation of water is 2257 kJ/kg, latent heat of fusion is 334 kJ/kg and if the pressure remains a constant 1 atm, determine the variation of entropy of the universe."

First of all, he hasn't covered first-order phrase transitions, so he can't use Clapeyron's equations (or the Clausis equation). The only thing I can think of is to use the liquid water generated from the heat of the vapor to obtain the specific internal energy of the mixture. However, since they say that the final temperature of the mixture is 0°C, I assume that means that atleast some ice remains, so why do they give the evaporation latent heat? What about the vapor, would that also have to be included in any entropy or energy balance? We barely have any data on it, just its initial temperature.

What do you guys think? Is all you need already included, any additional assumptions required?

Any help would be appreciated.
 

Answers and Replies

  • #2
nasu
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The system has the same temperature at the end and this is the solid-liquid equilibrium temperature. Then the final state will consist in liquid and solid water (ice). The heat extracted from the vapor is used to melt ice. So the extra water (amount given) is the mass of the vapor plus the mass of melted ice. And the heats need to balance. So you have two conditions and two unknowns (the mass of meted ice, mass of vapor).
 
  • #3
Andrew Mason
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Hey, all.

I'm tutoring a freshman, and we've come to this particular problem, and I think there's missing data. He does have access to the vapor tables, but as you can see, he's not given total mass, volume, nor any specific heat... here goes:

"An adiabatic calorimeter contains a water-ice mixture at 0°C and 1 atm. Vapor at 100° is introduced in the calorimeter until the mixture has gained 1.80 g of liquid water, the final temperature of the mixture being 0°C. If the latent heat of evaporation of water is 2257 kJ/kg, latent heat of fusion is 334 kJ/kg and if the pressure remains a constant 1 atm, determine the variation of entropy of the universe."
Just following up on Nasu's post, the ratio of water formed from condensation to water formed from melting should be in proportion to the specific heats. Together they add up to 1.8 grams.

Once you find these masses, it is just a matter of calculating the entropy loss of the steam to water at 100C, the entropy loss of the water from 100C to 0C, and the entropy gain of the ice at 0C and adding them up.

AM
 
  • #4
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Thanks, guys.

I can probably take over from there. ;)
 

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