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Drop ice in water, what is the final temperature of the water?

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data

    An ice cube, weighing 100g, is dropped in 1kg of water at 20 degrees C. Does the ice melt? If not, how much remains? What is the final temperature?

    The latent heat of fusion of ice at 0C is 6.025 kJ/mol, and the molar heat capacity of water is 75.3 J/K mol

    2. Relevant equations

    3. The attempt at a solution

    moles of H20= 55.51mol mol of Ice= 5.55mol

    heat contained in 1000g of water, using heat capacity, C=q/dT

    q= 75.3 J/K mol * 293.15K * 55.51mol= 1.225x10^6 J

    The heat required to melt 5.55mol of ice is:

    5.55mol x 6.025 kJ/mol= 3.34x10^4 J

    Here is where I get stuck. At first I tried to say that the heat of the water dropped by how much energy it took to melt the ice, and then use that as q and calculate T=q/C. This gives a reasonable answer, but then if you add enough ice, you would get an answer that is below 0C for final temp, which makes no sense.

    So what do I do now?

    EDIT: I think my above method is wrong because it neglects the energy of water from the melted ice. So if I calculate q for 1000g of water at 293K, then add that to q for water at 273K, and then divide that by C of water, should I get the right temperature?

    EDIT 2: Nope, I still get a temperature lower than freezing if I just add up all the heat of the system and divide by the heat capacity if I do the same calculation for 10x as much ice.
    Last edited: Oct 12, 2011
  2. jcsd
  3. Oct 12, 2011 #2
    The best way to go about this question is to write that the heat lost by the water is equal to the heat gained by the ice.
  4. Oct 12, 2011 #3
    to actually answer your question though, if you calculate the q for 1000g of water at 293K, then the q for 100g of ice at 273K, then add them together and subtract the heat that went into the phase change, you will be able to find the final temperature of all 1100g of water.
  5. Oct 12, 2011 #4
    That is what I did and I got temperature= 283.8K

    But if I do it again for 10x as much ice:

    heat of water = 1.225x10^6 J
    heat of ice= 1000g x 1mol/18.016g= 55.5mol x 75.3 J/K mol * 273 = 1140907 J
    heat put into phase change= 55.5mol x 6.025kJ/mol= 334.387kJ= 334387J

    Heat of water (at 293) + heat of ice (now turned to water at 273) - heat lost in phase change= 2031520 J

    T= q/C= 2031520 J/ [(55.5+55.5)mol *75.3J/K mol]=243.05K

    So the temperature is now almost -30C from putting in ice that is 0C...

    What is the issue here?
  6. Oct 12, 2011 #5


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    With your equations you are assuming that ice will melt at temperatures below 0oC. If ice does keep melting at temperatures below 0oC, you will indeed get temperatures below 0oC.

    Of course, because real ice obeys the laws of thermodynamics, it will not continue to melt once it reaches 0oC.

    Also, in your original calculation in the first post, you neglected to account for the energy required to warm the melted ice water (at 0oC) to the temperature of the rest of the water.
  7. Oct 12, 2011 #6
    Where is this issue represented in my equations?

    How do I do this? Is it possible to modify my previous post to include this measurement? I see my error in not including this, but I don't know what temperature it will warm to so I don't see how I can calculate it...

    So my equation now becomes:
    qtot= Heat of water (at 293) + heat of ice (now turned to water at 273) - heat lost in phase change - heat lost to raise temperature of ice

  8. Oct 12, 2011 #7
    Once the water gets to 0 degrees the ice won't melt. So you would have to add back in all of the latent heat from the phase changes that never happened.
  9. Oct 12, 2011 #8


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    This issue comes up because you assume that all the ice melts. This is not always the case.
    In thermochemistry problems such as this, it is often helpful to break the problem into smaller sub-steps. In the original problem, consider the following sub problems:

    1) When 100g of ice melts how much energy is absorbed?
    2) If the amount of energy in (1) is removed from 1kg of water at 20oC, what is the temperature of the water?
    3) When 100g of water at 0oC is mixed with 1kg of water at the temperature determined in (2), what is the final temperature?

    For problem #3, it is helpful to consider the final temperature as a variable (x), then write the heat required to heat 100g of water in terms of x, write the heat released when cooling 1kg of water to x in terms of x, then making these two expressions equal to each other.
  10. Oct 12, 2011 #9
    1) 6.025kJ/mol x 5.55mol= 3.34 x 104J

    2) qtot, water= 75.3 J/K mol * 55.5mol * 293.15K= 1225117 J

    1.23 x 106- 3.34 x 104= 1191679 J

    T= q/C = 1.23 x 106/ [75.3 J/K mol * 55.5mol]= 285.12 K

    3) So we have 1000g of water at 285.12 K and 100g of water at 273.15 K.

    Method one (weighted average):

    (100/1100)*273.15 + (1000/1100)*285.12= 284.03 K

    Method two (total heat of system):

    qH20 after phase change= 1191679 J
    qice after phase change= 5.55mol * 75.3 J/K mol * 273.15 K= 114153 J

    qtotal= 1191679 J + 114153 J = 1305832 J

    Temperature= qtot/[Cp,m* moltot]= 1305832 J / [61.05 mol* 75.3 J/K mol]= 284.06 K

    Both methods should give the answer, maybe the difference is due to rounding?

    Now if you have 1000g of ice and the same amount of water it takes 10x as much heat to melt the ice. But the water can only give 75.3J/K mol* 55.5mol* 20K= 83583J before it is at 0oC. This corresponds to 13.87 mol of ice, or 249.88g of ice. So there will be 750.12g of ice, and 1249.88g of water, all at 0oC.

    Let me know if I missed anything.

  11. Oct 14, 2011 #10


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    Yes, your calculations are correct (the small difference is due to rounding), and it looks like you understand everything well.

    Now, I have one more issue that I'd like to bring up. In many of your calculations, you calculate the amount of heat contained in the water by multiplying the amount of water with the heat capacity of water and its absolute temperature. This equations assumes that the amount of heat contained in the water is directly proportional to the temperature but this assumption is not true.

    Below is a diagram (not to scale) illustrating the temperature of water as more heat is added to the water (starting from absolute zero):
    As you can see, there are regions where the temperature of the water increases linearly with the amount of heat added. These regions correspond to the three phases of water: solid (blue region), liquid (green region), and gas (red region). At the boundaries between these regions, however, the temperature of the water remains constant even as more heat is added. These correspond to the phase changes (i.e. melting and vaporization) and the latent heats associated with these phase changes. In these regions, the added heat does not go into speeding up the particles to increase their temperature, but instead toward breaking up intermolecular interactions between water molecules.

    The diagram clearly demonstrates that the heat contained in a certain amount of water is not directly proportional to temperature. There are certain regions where this assumption holds true (for example, in the region between 273 and 373 K), but once you move outside these regions, the assumption breaks down.
    Last edited: Oct 14, 2011
  12. Oct 15, 2011 #11
    I understand what you are saying, and it leads me to wonder why my method worked then.

    When I calculated the final temperature of the water I, like you said, assumed that all the heat contained in the water was just a function of it's temperature, i.e. q=C*T. But really, if you start with water at 0K and bring it up to 293K, qtot=C*T + qfus

    So when I just used qtot= C*T, and then subtracted the heat required to melt the ice, why did it give me a correct final temperature? I assume it is because in correcting for my error, something else would be changed that would make the answer stay the same, but I am not sure what that would be.

    EDIT: Actually, does it have something to do with the fact that since I am calculating the heat change for a temperature change only, with no phase change included, I can use q that is only involved in changing the temperature, i.e. just q=C*T?
    Last edited: Oct 15, 2011
  13. Oct 15, 2011 #12


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    Yes, the qtot values you calculate are all off by the same constant amount. However, because this constant stays the same between 273K and 373K and because your calculations only relate changes in heat to changes in temperature, the constant ends up getting canceled out.
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