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Putting ice into a calorimeter, checking my thoughts

  1. Nov 26, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Ok, this is something related to a real life problem, more precisely a laboratory experiment.
    Say that the temperature of the room is 25°C. I have a calorimeter. I chose to put 250 g of water at 27.5 °C inside it. But as it is not a perfect calorimeter, say I've calculated (by the least squares method and by doing a long experiment) its equivalent mass in water to be 50 g. In other words, the calorimeter behaves like absorbing the same heat as 50 g of water.

    Now I want to put a 20 g ice cube in order to calculate the latent heat of water.
    To find it, I think of using the formula [tex]Q_1+Q_2=0[/tex]. Or, in details : [tex](M+\pi)c_{\text{water}}(T_f-27.5)+mc_{\text{ice}}(0-T_i)+mc_{\text{water}}(T_f-0°C)+mL_{\text{ice}}=0[/tex] where [tex]T_i[/tex] is the initial temperature of the ice (which might be below 0°C), [tex]T_f[/tex] is the final temperature of the calorimeter and the melted ice cube, M is the initial mass of water, that is 250 g, [tex]\pi[/tex] is the equivalent mass in water of the calorimeter.

    After measuring [tex]T_f[/tex], the only unknown in the equation would be [tex]L_{\text{ice}}[/tex].


    Ideally, what I would like is to have a [tex]\Delta T[/tex] such that the room temperature ([tex]T_r[/tex]) is such as [tex]T_r-\frac{\Delta T}{2}<T_r<T_r+\frac{\Delta T}{2}[/tex].

    [tex]\Delta T[/tex] is the difference of temperature between the initial water+calorimeter temperature and the final temperature of the water+calorimeter+melted ice cube.

    So ideally, in my example I would like to have [tex]T_f=[/tex]25°C-2.5°C=22.5°C, although I know it will likely not happen in my example.

    I wrote all this to ask... am I making any error somewhere? I feel I'm right on everything, but I might miss something. Especially regarding the equation [tex](M+\pi)c_{\text{water}}(T_f-27.5)+mc_{\text{ice}}(0-T_i)+mc_{\text{water}}(T_f-0°C)+mL_{\text{ice}}=0[/tex].
    Thank you infinitely.
     
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  3. Nov 26, 2009 #2

    ehild

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    You can use the equation of heat balance for a process only in a closed system: that is why one uses a calorimeter. And the calorimeter must be isolated from its surrounding so that the heat exchange with the room is negligible during the ice melting process. So the room temperature is irrelevant in your problem, you need to get a final temperature which is lower than 27.5 °C.

    ehild
     
  4. Nov 27, 2009 #3

    fluidistic

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    Thank you for helping.
    I can consider the calorimeter to be a closed system, that's why I calculated its equivalent mass in water. In order to do so I had to put a colder than room temperature water inside it, measure each 30 s the temperature of the water, for more than 10 minutes. After this, I put hot water compared to the room temperature and I measured the temperature of the cold+hot water each 30 s for more than 10 minutes. The cold+hot temperature was chosen so that the final temperature is still over the room temperature. I used the least squares method to calculate [tex]\Delta T[/tex].
    I get that [tex]\pi =\frac{m(T_{\text{hot water}}-T_f)}{\Delta T_{\text{corrected by least squares}}} -M[/tex]. The room temperature was relevant in the calculation of [tex]\pi[/tex] and also for calculating [tex]L_\text{ice}[/tex].
    Let me explain you why : say I've chosen to put water at 80°C inside the calorimeter. I chose to put the ice in it. After a very short time the ice completely melt and I can read 66°C. The temperature will still continue to decrease until reaching the room temperature, and it is totally impossible to me to determine the term [tex]T_f[/tex] in "[tex]mc_{\text{water}}(T_f-0°C)[/tex]" in the first equation of first post I gave.

    I realize it seems a bit contradictory, I consider the calorimeter to be perfect since I took in count how much heat it absorbs, but then I also care about how the temperature of the water inside it changes with time, according to the room temperature.

    I attach a file which represents the experimental data one of the many many experiments I did to calculate [tex]\pi[/tex].
     

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  5. Nov 27, 2009 #4

    ehild

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    Well, it is a difficult problem, if you work with a calorimeter which is not isolated from the ambient enough. Still, I think that the room temperature is irrelevant in your task, to get the latent heat of ice. I do not quite understand what you did, but I am afraid, you can not obtain the heat capacity of the calorimeter in the way you did. If you pour either cold water or hot water into the calorimeter, there are two heat exchange processes: a fast one, exchange of heat between calorimeter and water, the other is a slow one: heat exchange of calorimeter/water system with the surrounding air and room. This process is exponential in time and is described by Newton's law of cooling. During this process, the final temperature is that of the room. You can determine the heat capacity of the calorimeter from the fast process.

    You can use the calorimeter in such experiments which happen fast enough, so the heat exchange during this short time with the ambient is negligible. I think, you can stir the liquid in the calorimeter. Stirring makes the heat exchange faster.

    How can you determine the heat capacity ( [itex] \pi [/itex] ) of the calorimeter?

    Start with the state when the calorimeter has been long enough at room temperature. Read this initial temperature of the calorimeter, Tc(i). Warm water of a given volume V, measure the temperature of the water Tw(i), then pour the hot water into the calorimeter, close the calorimeter, start to measure time and temperature. At the beginning, the temperature decreases fast then it reaches an almost steady value, which will decrease further, but very slowly, compared to the speed of change during the first seconds. You can extend the slow linear part of the temperature-time plot to zero time, and consider this temperature as the final common temperature of water and calorimeter.
    Now you know the initial temperature both water and calorimeter, you know the common temperature and you know the mass of water. From that , you get the heat capacity of the calorimeter.

    Of course, you can do the same experiment with cold water.

    When you do the experiment with ice, you can follow the same method. Poor water into the calorimeter, let it stand long enough to take room temperature. Add the piece of ice. Or you can choose using hot water, but then measure the temperature of water before you pour it into the calorimeter and add the ice fast. Close the calorimeter, stir the water, trace the temperature and if you detect steady change of temperature, determine the final temperature of the melting process by extrapolating to zero time.

    You said that the piece of ice melted very fast and the temperature was 66 °C, when you used 80°C water. You can take this temperature as the final one, it must be close to the temperature you would get by extrapolating to zero time.

    ehild
     
  6. Nov 27, 2009 #5

    fluidistic

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    I appreciate your time and effort.
    [tex]\pi[/tex] is not the heat capacity of the calorimeter but it is the mass of water that would absorb or release heat as much as the whole calorimeter. So that if I put 200 g of water into the calorimeter and it stabilizes at say 23°C, I consider a perfect calorimeter that contains [tex](200+\pi)[/tex] g of water at 23°C.

    You are absolutely right, when the calorimeter is at a different temperature from its environment, it exchanges heat according to Newton's law of cooling. To minimize this effect, we chose a [tex]\Delta T[/tex] quite small (around 4°C). So for instance if the room temperature is 20°C, I put water at 22°C. The ice would decreases the temperature at around 18°C and the difference of temperature between the calorimeter and the room temperature is relatively small.

    Look at my new attached file : after 10 minutes, the calorimeter has lost around 0.1°C, which is almost nothing. Furthermore I take this into account and I do a correction with the least square method. The graph represent the temperature of the water+calorimeter at initially 26.2°C, after around 13 minutes I put 20 g of ice into the calorimeter and it reaches 21.1°C as a minimum. I obtained a [tex]L_{\text{ice}}=(78 \pm 4)\frac{cal}{g}[/tex] which is similar to the "real" one. The room temperature is 24.1°C.
    I don't know if I'm repeating, but I chose to put water a bit hotter than the room temperature so that the ice put it a bit colder than the room temperature so that I can read graphically when exactly the temperature of the calorimeter+ice+water reaches its minimum, because when it reaches it, it then starts to increase to the room temperature. But you said that even if the room temperature was below the minimum of the calorimeter+water+melted ice I could still read, quoting you in order not to be lost in the thread :
    , The big problem is that I can't see inside the calorimeter : I can't know when the ice has melted. Using my method, although not perfect at all, I can get the [tex]\Delta T[/tex] graphically, or via a correction in order to get a more accurate result.

    In fact I was doubting about my first equation. In the lab report, many students considered the ice to be at 0°C, hence the term [tex]mc_{\text{ice}}(0-T_i)[/tex] does not appear in my first equation. Professors put them OK for the equation.
    Take note that English is not my mother tongue and I'm more than sure that I'm not explaining all as I should. There are many parts I didn't talk about, like an estimation of [tex]\pi[/tex] in order to calculate at what temperature I should warm the water. What I did exactly with the least squares method, etc.
    The bibliography of the experiment comes from I book in Spanish which is very old and doesn't seem to have been translated. I can't even take it out of the library at my University. They keep an old photocopy of it and don't allow anyone to leave the library with this photocopy. So I don't have a good access to it. I could only read some parts of it in front of the librarian, I couldn't really study from it.

    P.S.: I constantly stirred the ice into the water with something I took into account when calculating [tex]\pi[/tex].
     

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  7. Nov 27, 2009 #6

    ehild

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    I think you performed a very careful experiment.

    The initial temperature of ice should be very close to zero if you keep it in a thermos for a while so some of it melts, but you can check it with a thermometer.

    The water equivalent of the calorimeter and its heat capacity differ only in the factor of specific heat of water. But I still do not understand how did you get it. What is deltaT, M, m and what is the final temperature in that experiment.

    Choosing the initial and final temperature is not so vital in the ice-melting experiment. If you succeeded to perform it nearly around the room temperature, it is enough.

    All what you have done with the ice-melting experiment seems correct to me, and your result is excellent for a student's lab measurement. I see that you knew and applied the method of "extrapolating to zero time" I tried to suggested.

    Do not worry about your English, it is very good, better than my one (I am Hungarian).

    I hope you get a very good mark !
     
  8. Nov 27, 2009 #7

    fluidistic

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    Thank you for all!
    M is the mass of the water in the calorimeter. [tex]\Delta T[/tex] is the measured difference of temperature between the mass of water+calorimeter just before putting the ice and the minimum of temperature the calorimeter+water+melted ice reaches.
    m is the mass of ice.
    Why applying the least squares method to determine a corrected [tex]\Delta T[/tex], call it [tex]\Delta T'[/tex], I had to trace some straight lines on the graph which gave me a corrected [tex]\Delta T'[/tex] that I used in the formula for pi : [tex]\pi =\frac{m(T_{\text{hot water}}-T_f)}{\Delta T_{\text{corrected by least squares}}} -M=\frac{m(T_{\text{hot water}}-T_f)}{\Delta T'}} -M[/tex]. I did exactly the same method in order to get [tex]L_{\text{ice}}=c_{\text{water}}\left [ \frac{(M+\pi)}{m}\Delta T' -T_{\text{minimum reached by calorimeter+water+melted ice}} \right ][/tex].

    The problem is that this [tex]L_{\text{ice}}[/tex] doesn't take into account that the ice can be below 0°C. As you said, I could check it out with a thermometer. To do so I would put 2 ice cubes together and compress a thermometer in between (a thermocouple). Hopefully it will give 0°C so that I won't have to add another term, but I don't think it will give 0°C.
    My final exam is on next Monday. There is maybe 50% of chance I'll get to do this experiment or 2 others very similar. (calculating the heat of vaporization of water or calculating the specific heat of iron).
    If you don't mind, I'll post here how I did. It is extremely important for me to do well.
     
  9. Nov 27, 2009 #8

    ehild

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    If you use ice taken out of the fridge, and you measure its mass and temperature, during this time the outside warms up, the inside remains cold....I think it is better to ensure 0 °C for the whole ice at the beginning. You can smash the ice, and keep it on air till it gets a bit wet. The water and ice is nearly in equilibrium then, hopefully, at 0 °C.

    Looking at the plot you sent, it took about 3-4 minutes to melt the piece of ice. It would be much faster with smashed ice. Rapidity is vital to ensure the conditions for a closed system. The heat exchange process between the calorimeter and surroundings is rather complicated, it is better to avoid this heat transfer as much as possible. That is why we use a calorimeter, instead of a simple glass flask. Your equipment is rather poor, but it is a student's lab. It is good if the student sees the problems connected with the measurement.

    Yes, post your other experiments. I will read.

    ehild
     
    Last edited: Nov 28, 2009
  10. Dec 1, 2009 #9

    fluidistic

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    Thank you for this reply. I agree the equipment was quite poor. But this was the idea of the numerous mathematical corrections we had to use and also to learn that the real problems aren't the ones in Resnick-Halliday although I don't criticize their fantastic book.

    As promised, I tell you my grade : 7/10. I had to work out the specific heat of a metal cylinder. It seemed copper to me and I obtained 0.10 cal/(g°C) while the one of copper is 0.09 cal/(g°C).
    I thank you for the idea of the smashed ice, I would never have thought about it.
    The calorimeter was a aluminum cylinder put in a wood can (hmm, here my English is probably too weak to explain well) with a wood top with 3 holes, to put the melter and the thermocouple and another hole for another purpose. Hence it loses heat easily as you can imagine.
    Glad to have talked to you, see you and take care.
     
  11. Dec 1, 2009 #10

    ehild

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    You are welcome.

    ehild
     
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