# Homework Help: Equilibrium temperature of some ice and steam

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1. Jan 18, 2017

### ElPimiento

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. Jan 18, 2017

### haruspex

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).

3. Jan 18, 2017

### ElPimiento

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.

4. Jan 18, 2017

### haruspex

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.

5. Jan 18, 2017

### ElPimiento

I'm going to bring it to my professor's attention. Thanks for the second opinion!

6. Jan 18, 2017

### CWatters

I got 23.15 C but perhaps I made an error somewhere.

7. Jan 18, 2017

### haruspex

No, I think you may be right. I suspect a sign error in the handling the f the latent heats. Will check.

8. Jan 18, 2017

### haruspex

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..

9. Jan 21, 2017

### CWatters

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.

10. Jan 21, 2017

### haruspex

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