How Does Entropy Change in a Water-Ice Bath with Vapor Introduction?

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In summary, Nasu thinks that there might be missing data, and that the entropy of the universe can be determined based on the initial temperature and the specific heats of water and vapor. However, he needs more data to confirm this.
  • #1
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.
 
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  • #2
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
Metaleer said:
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
Thanks, guys.

I can probably take over from there. ;)
 
  • #5


Hi there,

It seems like there is indeed some missing data in this problem. Without knowing the total mass, volume, or specific heat of the mixture, it is difficult to accurately determine the variation of entropy of the universe.

One approach you could take is to make some assumptions and use the given data to calculate the change in entropy. For example, you could assume that the initial mass and volume of the mixture is equal to the final mass and volume (since the final temperature is the same and the pressure is constant). From there, you can use the specific heat of water and ice to calculate the change in internal energy of the mixture, and then use the given latent heat values to determine the change in entropy. However, this approach would be limited and may not accurately reflect the true change in entropy of the universe.

Another approach would be to consider the entire system (water-ice mixture and vapor) as a closed system and use the first law of thermodynamics to calculate the change in internal energy. From there, you can use the given latent heat values to determine the change in entropy. This approach may be more accurate, but it would also require more assumptions and calculations.

In conclusion, it seems that additional data is needed to accurately determine the change in entropy of the universe in this scenario. I would recommend discussing this issue with your tutor and possibly reaching out to your instructor for clarification on the problem. Good luck!
 

What is the definition of entropy?

Entropy is a measure of the disorder or randomness of a system. It is a thermodynamic concept that describes the tendency of a system to move towards equilibrium.

How does entropy relate to water-ice baths?

In a water-ice bath, the ice is colder than the water and therefore, the water molecules have less energy. This leads to a decrease in the random movement of the molecules, resulting in a decrease in entropy.

Why is entropy important in science?

Entropy is important in science because it helps us understand and predict the behavior of systems. It is a fundamental concept in thermodynamics and is used to explain the flow of energy and matter in various processes.

How is entropy calculated?

The change in entropy (∆S) of a system is calculated using the equation ∆S = qrev/T, where qrev is the reversible heat transfer and T is the temperature in Kelvin.

What factors affect the entropy of a water-ice bath?

The entropy of a water-ice bath is affected by the temperature difference between the water and ice, the amount of ice present, and the amount of energy transferred into the system.

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