Solving Thermal Physics Questions: Q1 Calorimeter & Q2 Ice & Water

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The discussion focuses on solving two thermal physics problems involving heat transfer. In the first question, the user struggles to determine how much steam must condense to achieve a final temperature of 50 degrees Celsius in a system with a copper calorimeter and water. Key equations involve the heat gained by the water and calorimeter equating to the heat lost by the steam and water. For the second question, the user calculates the energy required to melt ice and the energy released by warmer water, ultimately questioning how the final temperature can be zero when not all ice melts. The responses clarify that thermal equilibrium dictates that some ice remains unmelted, keeping the system at 0 degrees Celsius.
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hi there guys I am doing a thermal physics question and its hard please help me and point me in the right directions please thanks.


question 1

A 50g copper calorimeter contains 250g of water at 20degrees celsius how much steam must be condensed into water if the final temperature is to reach 50degrees?

what i did i tried applying these equations
Qcold= -Qhot
MwCw(T-Tw) = MxCx(T-Tx)
i let mw=0.25g

now the tricky bit is the steam getting condensed to 50degrees i tried using the latent heat of evaporation but with no sucess and how does the copper insulator relate? pleaes help

Question 2
An insulated vessel 250g if ice is at 0degrees and is added to 600g of water at 18degrees a) what is the final temperature of water

ok first step here i calculated the energy required to change the phase of ice to water
Q=83.3 KJ

then i calcualted the energy 600g of water released and got Q=45.2 KJ

how do i go from here to get final temperature of the system? and the book says the answer is 0 how is that? please show me thanks
 
Physics news on Phys.org
1. Heat gained by 250gms of water from 20deg.C to 50 deg.C is equal to the latent heat loss by 'x' gms steam at 100deg.C + sensible heat loss by 'x' gms of water from 100deg.C to 50deg.C.

2. What will be the temperature of final mixture when you can't supply heat to melt the ice totally?
 
Hello,
1) I think that you'll have to use the relationship
Heat gained by copper calorimeter and water=Heat lost by Steam via condensation +Heat lost by 100 degree celcius of water(after condensation) to 50 degree celcius
2) For question 2, I think that you require the concept of thermal equilibrium. Since the energy released by the water is not enough to melt all the ice, it logically follows that some ice remains and the remaining water is also at 0 degree celsius as all of their heat ennergy is lost to melt the ice.
 
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