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Equilibrium temperature of some ice and steam

  1. Jan 18, 2017 #1
    1. The problem statement, all variables and given/known data
    "A well-insulated bucket of negligible heat capacity contains 120 g of ice at 0°C. If 20 g of steam at 100°C is injected into the bucket, what is the final equilibrium temperature of the system?"

    2. Relevant equations
    $$Q_{fus} = m_{water}L_{fus}$$
    $$Q_{vap} = m_{steam}L_{vap}$$
    $$Q = mc_{water}\Delta T$$
    Where ##L_{fus} = 334 \frac{kJ}{kg}##, ##L_{vap} = -2230 \frac{kJ}{kg}## (since the phase change is from gas to liquid), and ##c_{water} = 4.184 \frac{kJ}{kgK}##

    3. The attempt at a solution
    (I'm beginning to suspect the error is a computational one, or that I've given some term the wrong sign)

    Since the container is well-insulated I can write that no heat is lost:

    $$\begin{align*}
    \sum Q & = 0 \\
    Q_{fus} + m_1c_{water}\big(T_f - (273K)\big) + Q_{vap} + m_2c_{water}\big(T_f - (373K)\big) & = 0 \\
    \end{align*}$$

    Where ##m_1## refers to the ice and ##m_2## refers to the steam. Some algebra,

    $$\begin{align*}
    \big(m_1c_{water}T_f - m_1c_{water}(273K)\big) & \\
    + \big(m_2c_{water}T_f - m_2c_{water}(373K)\big) & = -(Q_{fus} + Q_{vap}) \\ \\
    \big(m_1c_{water}T_f + m_2c_{water}T_f\big) & \\
    - \big(m_1c_{water}(273K) + m_2c_{water}(373K)\big) & = -(Q_{fus} + Q_{vap}) \\ \\
    T_fc_{water}(m_1 + m_2) & = -(Q_{fus} + Q_{vap}) \\
    &\ \ \ \ + \big(m_1c_{water}(273K) + m_2c_{water}(373K)\big) \\ \\
    T_f & = \frac {-(Q_{fus} + Q_{vap}) + \big(m_1c_{water}(273K) + m_2c_{water}(373K)\big)} {c_{water}(m_1 + m_2)}.
    \end{align*}$$

    Numerically, this becomes,

    $$\begin{align*}
    T_f & \approx \frac {-(-4.52 kJ) + \big((137 kJ) + (31.2 kJ)\big)} {0.586 \frac{kJ}{K}} \\
    & \approx 294 K.
    \end{align*}$$

    or 21.7°C; but this is incorrect.

    Thanks,
    Andrew
     
  2. jcsd
  3. Jan 18, 2017 #2

    haruspex

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    I make it nearer 22. It looks like you used 273K converting to K but 272.7 converting back. Or maybe you accumulated some rounding errors.
    There's no need to work in K here, easier in C:
    ((4.52/4.184 + 100 * 0.020)/(0.120+0.020).
     
  4. Jan 18, 2017 #3
    I'm submitting the question online so here are the values I know are incorrect: 21.7°C, 21.8°C, 22°C, and 8.998°C.
    So, I hope it is not the case that the margin of error is crazy small, I also hope there is nothing fishy going on with the answer to the question that the website is using...

    Thanks for the suggestion about using C.
     
  5. Jan 18, 2017 #4

    haruspex

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    I got 22.00, so either the app is wrong or it is being very picky about significant digits. Maybe it wants 22.0, but I cannot see why that would be preferred to 22. The masses are not given as 20.0 etc.
     
  6. Jan 18, 2017 #5
    I'm going to bring it to my professor's attention. Thanks for the second opinion!
     
  7. Jan 18, 2017 #6

    CWatters

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    I got 23.15 C but perhaps I made an error somewhere.
     
  8. Jan 18, 2017 #7

    haruspex

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    No, I think you may be right. I suspect a sign error in the handling the f the latent heats. Will check.
     
  9. Jan 18, 2017 #8

    haruspex

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    Edit: no, I still get 22:
    Heat left over from conversion of all to water = 20x2230-120x334=4520J.
    (4520/4.184+100*120)/(120+20)=22.00218..
     
  10. Jan 21, 2017 #9

    CWatters

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    Think I might have found the problem. The calculation appears sensitive to the exact value used for the latent heat of vaporization.

    If I use the figure from here..
    https://en.wikipedia.org/wiki/Latent_heat
    ..which has the latent heat of vaporization as 2264.76 kJ/kg then I get a figure of around 23C.
     
  11. Jan 21, 2017 #10

    haruspex

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    Ah.. I didn't check those data. Online I see 2230, 2257, 2260, 2.3 x 103. 2230 is rather an outlier.
     
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