Proportion of metal in each alloy?

[Marshall878]

Once again I've left homework to the last minute so my teacher will moan if i go ask him tomorrow morning. I've scanned the net and all my textbooks but can't seem to find a working example that helps with what i am doing. Could someone look at these 2 questions and help me out?

1. An alloy has a density of 8000kg/m^3 (metres cubed). It is made from two metals of densities 10 000kg/m^3 and 7 200kg/m^3. What is the proportion of each metal (by mass) in the alloy?

2. 0.000 500m^3 of a liquid of density 800kg/m^3 are mixed with 0.000 300m^3 of a liquid of density 1 200kg/m^3. What is the density of the mixture, if there is no change in volume on mixing? (when adding two liquids of different densities together, they don't mixx do they?)

Much appreciated

Lee

 

[Diane_]

What have you done on the problems so far?

 

[Marshall878]

well i haven't done anything on them, i keep trying but nothing seems right. Weve been told the formula to use throughout these questions p=M/V but i don't know where to start, if someone could point me in the right direction i might be able to get it

with the liquid one, we haven't done anything to do with liquid and none of the texts book i have been given include anything about adding different density liquids together. What happens to the density because when two liquids of different densities are added together, they don't become one, they seperate.

 

[Diane_]

Well, try this:

1) Suppose you mixed one cubic meter of the first metal with one cubic meter of the second. What would the resulting density be? That should start pointing you in the right direction.

2) It doesn't matter whether they mix or not - you're just looking for the overall or average density. You know the volume of the result and you should be able to figure out the mass - that ought to do it for you.

 

[Marshall878]

geeze i feel really thick

would one cubic metre of the first metal just be 10,000 kg and then one cubic metre of the second metre would be 7,200kg?

so you add them together to give you 17,200kg. But the density of the alloy is only 8000kg/cubic metre.

 

[Diane_]

And how many cubic meters of it would you have?

 

[Marshall878]

17500/8000 = 2.1875 cubic metres?

 

[Diane_]

Marshall, you seem to be flailing about with this, and I'm not sure if you really don't understand or if you're just not putting the necessary thought into it. I'm not sure where the 17,500 came from, what the units on it are, or why you're dividing it by the density of one of the two metals. If you take a one cubic meter block of metal and combine it with another one cubic meter block of metal, how many cubic meters of metal do you then have? That would be your volume. You have the mass. You should be able to get there from here.

 

[Marshall878]

1. An alloy has a density of 8000kg/m^3 (metres cubed). It is made from two metals of densities 10 000kg/m^3 and 7 200kg/m^3. What is the proportion of each metal (by mass) in the alloy?

I don't have the mass, just the densities of each metal that combine to make the alloy.

No I am not going into this half heartidly, I've been sitting here since half 3 and its now nearly half 9, going through these 2 questions, and i keep comming back to the beginning, why did i take physics further than GCSE

 

[Werg22]

Question 1.
Let [tex]x[/tex] be the mass of the first metal and [tex]y[/tex] the mass of the seconde. And let [tex]a[/tex] be the volume of the first metal and [tex]b[/tex] the volume of the second. Now you know that the mass of the alloy is equivalent to [tex]x + y[/tex] and its volume is [tex]a + b[/tex]
Therefore

[tex]\frac{x+y}{a+b} = 8000 [/tex]

Also we know that

[tex]\frac{x}{a} = 10 000[/tex]

[tex]\frac{y}{b} = 7200[/tex]

Now the relation you are looking for is [tex]\frac{x}{y}[/tex]

So
[tex]\frac{x}{a} = 10 000[/tex]
[tex]\frac{x}{10 000} = a [/tex]
[tex]\frac{y}{b} = 7200[/tex]
[tex]\frac{y}{7200} = b[/tex]

Also

[tex]\frac{x+y}{a+b} = 8000 [/tex]

[tex]a + b = \frac{x+y}{8000}[/tex]

So

[tex]a + b = \frac{x+y}{8000}[/tex]

[tex]a + b = \frac{x+y}{8000}[/tex]

[tex]\frac{x}{10 000} + \frac{y}{7200} = \frac{x+y}{8000}[/tex]

[tex]\frac{x}{100} + \frac{y}{72} = \frac{x+y}{80}[/tex]

[tex]16x + 25y = 20x + 20y[/tex]

[tex]16x + 25y = 20x + 20y[/tex]

[tex] 4x = 5y[/tex]

[tex] x = 5/4 y[/tex]

If you want the proportion;

[tex]\frac{x}{y} = \frac{5/4 y}{y} = 5/4 [/tex]

So x/y = 5/4 and y/x = 4/5

Question two:

I think this is very easy after that... remember the volume will to sum of the volume is the volume of the mixture and the sum of the mass is the mass of the mixture.

 

[Diane_]

Marshall - You don't really need masses. You have the densities, and the mass is factored into that. One way to approach the problem is as I suggested earlier - try coming up with specific volumes and working with those. You'll need to do as Werg suggested, and eventually work directly with the data you have, but sometimes when you're unsure what to do, it helps to get specific and see how things work out there.

I know it can be frustrating. Everyone who's ever taken any math or science course beyond the absolute basics has felt that frustration. What you're learning to do here is deal with abstract concepts, and that's definitely a good thing.

Go back to your first problem, assume one cubic meter of each substance and determine the density of the result. When you see how that works, try using variables for the volumes - again, as Werg suggested. Remember that you're looking for the proportion, which means you'll be looking for the ratio of the two volumes. You will not be able to solve for the volumes directly - it's not that kind of problem. But you should be able to rearrange things algebraically to get the ratio.

And, if you haven't already done so, let me suggest that you not read through Werg's post in detail until you've made the attempt on your own. You can use it as a guide if you get stuck, but following it directly is like using a walkthrough in a computer game. You finish the game, but where's the fun in it?

 

[Marshall878]

Well I am tired now, but I've taken on board everything you two have suggested. I am going to go sleep on it and have a crack at it in the morning. I've printed off what both of you have said and will work upon it.

Thanks a lot and ill report back tomorrow with any results

**are these kind of questions common when starting an A-level course? they seem pretty intense or is it just me struggling with the jump from GCSE to A-level?**

 

[Werg22]

Remember that mathematic is another word for logic. Look at how the problem was analysed. Before doing anything on the paper think of a "plan". You have to figure out what you need to do in your head. This is a matter of practice and I suggest that you do a lot of problems.

 

[Diane_]

I'm afraid I'm unfamiliar with the difference between GCSE and A-level, hon - American, don't you know. :) However, it's a fairly basic question. As I indicated, I think the reason you're finding it difficult is that you're moving a step further into dealing with abstractions. That's a difficult step, and always a cause of frustration initially. But I can tell you that, when you finally master it (and you will), it can be a source of great pleasure. I also believe that all major human advances have come from people who were able to deal with abstractions successfully.

And I know that's probably cold comfort at this point. About all I can tell you is what my mother used to tell me when I got frustrated with something: Do it because it'll be good for you. That never convinced me, either, but it turns out she was right.