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Water poured over ice until temp is at equilibrium

  1. Jan 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose 337g of water at 32.9°C is poured over a 59.3g cube of ice with a temperature of -5.67°C. Does all the ice melt? If all the ice melts, what is the final temperature of the water? If all of the ice does not melt, how much ice remains when the ice-water mixture reaches equilibrium?

    Specific Heat of water = 4.19*103J/(kg*K)
    Specific Heat of ice = 2.06*103J/(kg*K)
    Latent Heat of fusion of ice = 334*103J/kg

    2. Relevant equations
    Q=mCΔT


    3. The attempt at a solution
    To determine if the process results in either ice, or ice-water mixture, I need to calculate how much heat is required to turn all the ice into water, then calculate how much heat was actually given off by pouring the water onto the ice.

    So first

    Heat required to raise the temperature of ice, and melt it completely
    Q1=miceCiceΔT+miceLheat of fusion

    Q1=(0.0593kg)(2.06*103J/(kg*K))(5.67°C)+(0.0593kg)(334*103)J/kg)
    Q1=20498.84J

    20498.84 Joules required to raise the temp of the ice and melt it completely.

    Now how much heat is given off when decreasing the temperature of the water to 0°C
    Q2=mwaterCwaterΔT

    Q2=(0.337kg)(4.19*103J/(kg*K))(32.9°C)
    Q2=46455.79J

    More heat is required to freeze the water than to melt the ice, so all the ice will melt.

    Here is where I am stuck. So I need to find the final temperature of the water at equilibrium.
    To melt the ice requires 20498.84 Joules so I set it equal to mwaterCwater(Tf-Ti) and solve for Tf.

    This would tell me the temperature of the water right after the last bit of ice melted into water.

    Rearranging this gives me

    [Q1+mwaterCwaterTi]/[mwater]Cwater = Tf

    Tf=65.8°C (which is impossible since it can't get hotter than when it started)

    I should be able to use Tf as the initial temperature of the next phase (rising temp of total water) heading towards equilibrium.

    Am I not accounting for something? Should I not set
    Q1=mwaterCwaterΔT
    to solve for Tfinal

    Even after I do that, all I know is the temperature of the total water after the ice melted but have no way of knowing the final temp at equilibrium. If I want to use Q=mCΔT i have two unknows, Tfinal and Q, since I don't know how much heat is taken in as the mixture goes from colder freshly-melted ice-water + warmer liquid water to equilibrium.

    What am I not understanding conceptually?

    Thanks in advance.

    P.S. Don't be rude please.
     
    Last edited: Jan 23, 2014
  2. jcsd
  3. Jan 24, 2014 #2

    SteamKing

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    I follow your calculation of the heat required to melt the ice completely and the amount of heat available in the water which is poured over the ice.

    Remember, once the ice has melted to liquid water, its mass mixes with that of the water being poured. You know how much heat is required for melting the ice. You now have to figure out the balance of heat afterwards.

    You need to write an equation like this:

    Q(melt water) gained = Q(pouring water) lost
     
  4. Jan 24, 2014 #3
    I am not sure how you rearranged it:

    mwaterCwater(Tf-Ti)=Q1

    (Tf-Ti)=Q1/mwaterCwater

    With this you get the amount of kelvins, which the temperature of the water has decreased with (Tf-Ti) in order to melt the ice. You can then again use conservation of energy to do as suggested above, and remember to substract what you just found from the initial temperature of the water, and you should be able to find the equilibrium temperature.
     
  5. Jan 24, 2014 #4
    So by Qmelt water you are referring to 20498.84 J right? Which is equal to the amount of heat the poured water lost when it melted the ice completely.

    It should look like this
    Qmelt water=mpoured waterCwater(Tf-Ti)
    correct?

    The only value here that I don't know is the Tf so solving for it my equation becomes
    Qmelt water=mpoured waterCwaterTf-mpoured waterCwaterTi

    Qmelt water+mpoured waterCwaterTi=mpoured waterCwaterTf

    (Qmelt water+mpoured waterCwaterTi)/mpoured waterCwater=Tf

    Now I plug in the values and solve for Tf

    [(20498.84J)+(0.337kg)(4.19*103J/(kg*K))(305.9K)]/[(0.337kg)(4.19*103J/(kg*K))]=Tf

    (20498.84J+431939.98J)/1412.03J/K)=Tf

    320.417K=Tf

    but 320.417K=47.4°C

    The water started at 32.9°C, it can't possibly have gotten hotter. I know the math is correct so I am not using the correct information.

    Am I correct to assume that for this last calculation, the Tf should be the temperature of the mixture when ice melts completely, and it is only at this time when I should consider the TOTAL mass of melted ice and poured water, and I should use this value to then determine that the Tf is as it goes from mixture of (poured water)+(freshly melted ice) to total water at equilibrium which is the next phase of this process?
     
    Last edited: Jan 24, 2014
  6. Jan 24, 2014 #5
    Doing this my Tf came out to 320.41K

    Which is higher than when it started. This is not possible.

    Tf-Ti=(20498.84J)/(0.337kg)*(4.190*103kJ/(kg*K))

    Tf-Ti=14.517K

    Tf=14.517K+Ti

    Tf=14.517K+305.9K=320.41K

    So instead I used the mass of the total water (melted ice cube+poured water)...

    Tf-Ti=(20498.84J)/(0.3963kg)*(4.190*103kJ/(kg*K))

    Tf-Ti=12.35K

    Tf=12.35K+Ti

    Tf=12.35K+305.9K=318.24K

    Still that temperature is higher which doesn't make sense.

    Am I skipping something? Am I supposed to consider the heat Q1 as negative? It is the only value that could possibly be negative to result in a decrease of the initial temperature. If I am cooling the water that began at 305.9K then is the heat considered negative because it is being lost?

    Thanks
     
  7. Jan 24, 2014 #6
    Well what I meant, was to calculate ΔT and substract it from the initial. For some reason I was unsure, which was your Tf and Ti.. But if you calculate ΔT = Q1/mwcw you get 14.5K, then substract that from the starting temperature, as you know this value has to be negative, as the water transfers energy to the ice cube. So you get Tf = Ti-14.5K

    But yeah a faster way is to just change the sign inside the paranthesis, since the energy has to be negative. You put Q1 = mwcwΔT but this is wrong, because Q1 has to be negative, when you are calculating it for the water, as it is the energy transfered from the water to the ice cube in order to melt it.

    Else you can't have conservation of energy, and the 1st law is disobeyed.
     
  8. Jan 24, 2014 #7
    Ok makes sense. So my final temperature is 305.9K-14.5K=291.38K when the ice has fully melted. The mixture however has not necessarily reached equilibrium.

    Now the last part should be the part of the process that goes from 291.38K to equilibrium correct? We shall call it, let's say, Q3

    Q3=mtotal waterCwater(Tf-Ti)

    But here I have two unknowns, Q3 and Tf

    Up to this point 20498.84 Joules have been lost by the poured water as it gets colder, but since it is an open system, more heat is lost to the surroundings as the mixture cools from 291.38K to equilibrium. Since this is the case, I don't have a specific value for heat that I can equate to Q3 so that I can solve the above problem. How can I know what amount to use for Q to be able to solve Tequilibrium?
     
  9. Jan 24, 2014 #8
    I think SteamKing already answered this last question of yours, and if you think more about it, it's quite obvious.

    You are now left with two substances of water, one at 291K and one at 273K, they have different mass, but same heat capacity. So you want to find an equilibrium temperature between each quantity, you call it Q3, which is fine.

    Now remember, that there has to be conservation of energy. The energy transfered from the water at 291K has to be equal to energy absorbed by the water at 273K, for both fluids to be in equilibrium.

    Try to set up two equations for Q3 and put them equal to each other, and see if it can then be solved, one for each quantity of water.
     
  10. Jan 24, 2014 #9
    Ok I see. So I would do

    Q3=mpoured waterCwater(Tf-Ti)

    should be equal to (due to conservation of energy)

    Q3=mmelted iceCwater(Tf-Ti)

    I should set them equal to each other and solve for Tf.

    mpoured waterCwater(Tf-Ti)=mmelted iceCwater(Tf-Ti)

    I can cancel out the heat capacity since it is on both sides...

    multiplying out gives me this

    mpoured waterTf-mpoured waterTi=mmelted iceTf-mmelted iceTi

    Solving for Tf

    mpoured waterTf-mmelted iceTf=mpoured waterTi-mmelted iceTi

    Factor the Tf out

    Tf(mpoured water-mmelted ice)=mpoured waterTi-mmelted iceTi

    Finally

    Tf=[(mpoured waterTi-mmelted iceTi)/(mpoured water-mmelted ice)]

    That should give me the correct answer. I see how conservation of energy works here, but is this a correct translation of the concept into calculation of it?
     
  11. Jan 24, 2014 #10

    SteamKing

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    I should have been more clear in my original post.

    The Q = 20498.84 J has already been used to melt the ice and turn it to liquid water at 0 C.

    There is an unknown final temperature Tf where the melt water and the poured water reach thermal equilibrium. It is for this circumstance that you write the equation:

    Q(gained by melt water) = Q(lost by poured water)
     
  12. Jan 24, 2014 #11
    Ok thanks yes I understood now what you were trying to tell me, so the calculations in my previous post are still correct right? I only needed the Q = 20498.84 J to determine the change in temperature of the poured water at 32.9 °C to allow me to calculate Q3 as jhelmgart explained?
     
  13. Jan 24, 2014 #12
    Well I might not have expressed myself clearly before. You did the math correctly, but remember that when a substance transfers heat (or loses heat) it has to be a negative value.
     
  14. Jan 24, 2014 #13
    So in my last post I have arrived at the final temperature correctly? Just double checking before I fully submit my assignment to check my score.

    Thanks again for the help. Much appreciated
     
  15. Jan 24, 2014 #14
    I don't think so. You have to do put a minus sign in front of Qpoured, or else they wont be equal. I think it would change the sign in the upper and lower bracket of your paranthesis?

    You would probably get a temperature that doesn't make sense the way you did it?
     
  16. Jan 24, 2014 #15
    ok so what I should have done was...

    -mpoured water(Tf-Ti)=mmelted ice(Tf-Ti)

    So the Q for the poured water should be negative since that process involves losing heat(getting colder) and the melted ice is positive because it is gaining heat (getting warmer).
     
  17. Jan 24, 2014 #16
    the reason that is true is because

    To clear up a conceptual misunderstanding what about the fact that

    Qtotal=Qice+Qwater

    Qice will increase in temperature and it will take in heat. This value will be positive.

    Qwater will decrease in temperature and it will lose heat. This value will be negative.

    If they reach equilibrium, then the magnitudes will be equal to each other.

    Qtotal=Qice+Qwater=0

    This means that 0=(a positive x value)+(a negative x value)

    If 0=(x)+(-x)

    x=x

    How could I make sense of a negative x value being equal to a positive x value?
     
  18. Jan 24, 2014 #17
    But Qwater itself, when you calculate it, gives a negative value, because the temperature Tf is smaller than Ti. This isn't the case for Qice.

    So they wouldn't be equal to each other unless you put the negative sign in front of Qwater.

    In other words Qtot = Qice + Qwater = 0, you then isolate it to be Qice = -Qwater, hence the negative sign.

    Perhaps I was a bit unclear or said something wrong. Anyway this reasoning is probably better :) good luck
     
  19. Jan 24, 2014 #18
    Got it! Thanks!
     
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