Mixing water and ice

[hmparticle9]

Homework Statement

I mix 500g of water at 20c and 200g of ice at 0c, what is the end product?

Relevant Equations

$$\Delta Q = mc\Delta T$$

The total change in heat is zero and we have to account for the ice melting. I came up with the following equation:

$$0 = 0.5 c_w (T_f - 20) + 0.2L +0.2 c_w T_f$$

Hence

$$T_f = \frac{10c_w - \frac{1}{5} L}{\frac{7}{10} c_w}$$

$L$ is the latent heat for melting the ice and $c_w$ is the specific heat of water. We have to melt the ice (the $0.2L$ term) and the other terms correspond to changes in the temperature of the other components of the mixture.

The next part is throwing me. In the back of my book, some constants are provided.

$$L = 334 \frac{\text{kJ}}{\text{kg}} \text{ and }, c_w = 4,184\frac{\text{J}}{\text{kgK}} = 4.184\frac{\text{kJ}}{\text{kgK}}$$

Putting these in get $T_f = 14$.

In the book the solution is we end up with 75g of ice and 625g of water at 0c.

They obviously do not agree. I think I should be getting $T_f =0$ and then I have to figure out how much ice actually melts. Help is appreciated.

 

[phyzguy]

How much heat has to be removed to drop the temperature of 500g of water by 20C? How much ice has to melt to remove that much heat?

 

[mjc123]

Putting these in get Tf=14.

No it doesn't; it gives a negative number. If c_w >> L, then T_f = 100/7 = 14.28. So I suspect you used 4,184 instead of 4.184 for c_w.

 

[hmparticle9]

Oh crumbs. Yes I get -8.52. But that really does not make sense. The final temperature is below both of the original ingredients. What does this mean? I have assumed that the ice would have melted, maybe there was not enough heat to melt the ice and that is why I am getting silly answers...

By $\Delta Q = mc \Delta T$, the removal of heat needed to drop 500g of water by 20C is 41.84kJ.

But then, using $\Delta Q = ML$, I obtain $M = 0.125$. That is, we have enough heat to cool the water down to 0C, and we have $0.2-0.125 = 0.075\text{kg}$ of ice remaining. The total quantity of water is $0.5+0.125 \text{kg}$

Thanks guys :)

 

[hmparticle9]

I have another problem like this, but I am getting a slightly different solution to the problem. Could I have some comments on my solution please?

I start off with 20g of steam at 100C and 150g of ice at 0C.

I have said that to melt the ice we need to put in 0.15 x 334 = 50.1 kJ of heat and to condense the steam take out 0.02 x 2257 = 45.14 kJ.

Hence we need to look for solution $T_f$ of

$$50.1 - 45.14 + 0.02 c_v (T_f - 100) + 0.15 c_v T_f = 0 \implies T_f = 4.79$$

The answer is the back is 4.7.

 

[mjc123]

Oh crumbs. Yes I get -8.52. But that really does not make sense. The final temperature is below both of the original ingredients. What does this mean? I have assumed that the ice would have melted, maybe there was not enough heat to melt the ice and that is why I am getting silly answers...

Yes, exactly. The nonsensical result means that the assumption that all the ice melts is wrong.
For a problem like this, the best way to start is by calculating 2 quantities:
(i) The amount of heat released by lowering the temperature of all the water to 0^oC;
(ii) The amount of heat required to melt all the ice.
If (i) is greater than (ii), you will end up with water at some temperature above freezing. If it is less, you will end up with a water/ice mixture at 0^oC. Set up appropriate equations in each case.

 

[kuruman]

I have another problem like this, but I am getting a slightly different solution to the problem. Could I have some comments on my solution please?

I start off with 20g of steam at 100C and 150g of ice at 0C.

I have said that to melt the ice we need to put in 0.15 x 334 = 50.1 kJ of heat and to condense the steam take out 0.02 x 2257 = 45.14 kJ.

Hence we need to look for solution $T_f$ of

$$50.1 - 45.14 + 0.02 c_v (T_f - 100) + 0.15 c_v T_f = 0 \implies T_f = 4.79$$

The answer is the back is 4.7.

For future reference, next time you post a problem "like this" please post on a separate thread otherwise there will be confusion as to what reply belongs to what problem. The difference between your answer and the book's is probably due to roundoff errors. Get a symbolic solution without numbers in the form $T_f=\dots$ and substitute numbers at the very end. A spreadsheet is wonderful for this sort of thing.

 

[hmparticle9]

I am sorry. I thought a separate post would be a bit much considering the problems are nearly identical. I will make sure to abide by the rules next time. :)

 

[kuruman]

The answer is the back is 4.7.

I did a 4 sig-fig spreadsheet calculation with values from the web
c_v = 4.186 J/(g⋅°C); L_fusion = 334 J/g; L_vapor. = 2256.4 J/g
and your values
m_ice=150.0 g; m_steam = 20.00 g.

For the final temperature these numbers gave me T_f = 4.778 °C.

The calculation is sensitive to the values of the latent heats. Just reducing the latent heat of vaporization by 0.15% to L_vapor. = 2253.0 J/g gave me T_f = 4.682 °C which is a 2% reduction of T_f.

Lesson to be learned: Don't do intermediate numerical calculations and don't round off. Derive the answer in symbolic form and substitute at the very end. Use a spreadsheet, if you can.

 

[Steve4Physics]

@hmparticle9, I also favour the use of spreadsheets where possible. But if unable to use one (e.g. under examination conditions) don’t forget calculators have memories which can be used to accurately store intermediate results.

Also, note that in the Post #5 question, data are given to only one or two significant figures. Strictly speaking, the final answer should be rounded to one sig. fig. but that seems a tad over-zealous. Two significant figures seems reasonable, so your final answer (which you have correctly calculated) would be rounded to 4.8 $^{\circ}$C.

 

[kuruman]

But if unable to use one (e.g. under examination conditions) don’t forget calculators have memories which can be used to accurately store intermediate results.

When I gave exams, I required no numerical answers and I banned calculators during the exam. I expected all answers to be expressions in terms of given quantities symbols for which were specified in the statement of the problem. This was a time-saving measure: (a) it saved students' time especially those who, instead of thinking, frantically push the buttons of their calculators looking for an answer that isn't there; (b) it saved me grading time because the algebra was transparent and made the assessment where and how seriously a student went wrong easy to ascertain.

 

[Steve4Physics]

When I gave exams, I required no numerical answers and I banned calculators during the exam. I expected all answers to be expressions in terms of given quantities symbols for which were specified in the statement of the problem. This was a time-saving measure: (a) it saved students' time especially those who, instead of thinking, frantically push the buttons of their calculators looking for an answer that isn't there; (b) it saved me grading time because the algebra was transparent and made the assessment where and how seriously a student went wrong easy to ascertain.

That’s a good approach. When I was teaching I was preparing students for external examinations set by national/international examination boards (mainly AQA and Edexcel). Examination questions typically required numerical solutions. Not ideal but very common and you had to adapt to it.

Preparing students to deal with this type of question under examination conditions was important. I suspect many of the posters here are preparing for such examinations.

 

[Herman Trivilino]

What does this mean?

It means you assumed all the ice melted.