Calorimetry Problem: Finding Final Temperature and Phase Change

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The discussion revolves around a calorimetry problem involving a copper calorimeter containing water and a cold copper body. The first part of the problem was solved correctly, determining the final temperature to be approximately 1.2°C. The second part raises questions about potential phase changes of water into ice when the copper body is at -50°C, with participants emphasizing the need to consider latent heat and heat transfer principles. It is suggested that a trial solution approach can help determine whether ice forms based on the calculated final temperature. Overall, the complexity of the problem highlights the importance of understanding heat exchange and phase transitions in calorimetry.
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Homework Statement


In a copper calorimeter of mass m=100g, there is water of mass m1=200g with a temperature of t1=4°C. A copper body of mass m2=300g and temperature t2= - 20°C is inserted into the calorimeter.
a) What will be the final temperature inside the calorimeter?
b) Show that one fraction of water would turn into ice if the inserted copper body were to have a temperature of t2= - 50 °C. Calculate the mass of the formed ice.

Homework Equations


Q=mcΔt and second law of thermodynamics

The Attempt at a Solution


To solve a) I wrote the equations for the heat that will leave the water+copper calorimeter system and for the heat that the inserted copper body will receive. Then I set them equal and found for what temperature are those two quantities the same (t=1,2°C). That is correct. For b) I don't have an idea how to solve it since I don't know how can I check whether there will be a phase transition of water during the process, and whether all water will turn into ice or just a portion. I don't either quite understand the solution of a), since I apply equations for calculating heat without phase transitions etc., but don't quite know how can I prove that there won't be a phase transition. Shouldn't that be the first step of the solution, and only then applying equation?
Also, I tried working with maximum heat that water can give to copper, but the problem is that I don't have any lower or upper value of temperature so I could calculate the maximum heat.
 
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CherryWine said:
I don't have an idea how to solve it since I don't know how can I check whether there will be a phase transition of water during the process, and whether all water will turn into ice or just a portion.

you know water forms to ice if it goes to temp 0 degree and you have the latent heat of fusion of ice- so calculate the heat transfer and see how much water will convert to ice.
CherryWine said:
I don't either quite understand the solution of a), since I apply equations for calculating heat without phase transitions etc.
you might have mixed hot water in the bathroom to a good volume of cold water -resulting in a warm bath water.
heat gets transferred from a body at higher temp. to colder body and one calculates the amount of heat transferred to raise the colder body to the mean temp and how much heat is transferred from hotter body to cool it down to mean or final temp.
so its like exchanging money such that both have equal money.
the hot one looses heat energy and cold one gains heat energy and both of the amount should be equal.
 
CherryWine said:
I don't know how can I check whether there will be a phase transition of water during the process, and whether all water will turn into ice or just a portion.
With these problems, you have to start with a trial solution, a guess at what mix of phases will exist at the end. If the temperature you calculate is consistent with that guess, you have the answer. If it is not, it will suggest which way to move to a new guess. E.g., if you guessed no ice but the temperature comes out below 0C then you know you guessed too high. Note that this does not immediately prove there will be ice... perhaps there is some other medium present with a transition temperature that would be crossed first.
In general, these problems can be quite complicated. For more info see section 3 of
https://www.physicsforums.com/insights/frequently-made-errors-heat-elementary-level/
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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