Equilibrium temperature of some ice and steam

In summary: I'll go with that then.Ah.. I didn't check those data. Online I see 2230, 2257, 2260, 2.3 x 103. 2230 is rather an outlier. I'll go with that then.
  • #1
ElPimiento
17
0

Homework Statement


"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?"

Homework 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}##

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
 
Physics news on Phys.org
  • #2
ElPimiento said:
21.7°C; but this is incorrect.
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
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
ElPimiento said:
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.
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
haruspex said:
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.
I'm going to bring it to my professor's attention. Thanks for the second opinion!
 
  • #6
I got 23.15 C but perhaps I made an error somewhere.
 
  • #7
CWatters said:
I got 23.15 C but perhaps I made an error somewhere.
No, I think you may be right. I suspect a sign error in the handling the f the latent heats. Will check.
 
  • #8
haruspex said:
No, I think you may be right. I suspect a sign error in the handling the f the latent heats. Will check.
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
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
CWatters said:
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.
Ah.. I didn't check those data. Online I see 2230, 2257, 2260, 2.3 x 103. 2230 is rather an outlier.
 

FAQ: Equilibrium temperature of some ice and steam

1. What is the equilibrium temperature of ice and steam?

The equilibrium temperature of ice and steam is 0 degrees Celsius, also known as the melting point of ice. This means that when ice and steam are in contact with each other, they will reach a temperature where both phases coexist in a stable state.

2. How is the equilibrium temperature of ice and steam determined?

The equilibrium temperature of ice and steam is determined by the energy exchange between the two phases. As heat is added or removed, the temperature will change until it reaches a point where the rate of energy transfer between the phases is equal, resulting in a stable equilibrium temperature.

3. Why is the equilibrium temperature of ice and steam important?

The equilibrium temperature of ice and steam is important because it is a fundamental concept in thermodynamics and is used to understand and predict the behavior of materials undergoing phase changes. It also has practical applications, such as in the design of refrigeration systems.

4. Does the equilibrium temperature of ice and steam change under different conditions?

Yes, the equilibrium temperature of ice and steam can change under different conditions, such as changes in pressure or the addition of impurities. For example, increasing the pressure can result in a higher equilibrium temperature, while the presence of impurities can lower the equilibrium temperature.

5. Can the equilibrium temperature of ice and steam be different for different types of ice or steam?

Yes, the equilibrium temperature of ice and steam can vary depending on the type of ice or steam present. This is because the equilibrium temperature is influenced by factors such as the molecular structure and energy content of the phases. For example, the equilibrium temperature of water ice will be different from the equilibrium temperature of dry ice (solid carbon dioxide).

Back
Top