Mass, density, volume: Find the volume of an alloy

[dominicwild]

Homework Statement

"An alloy of copper and tin has a volume of 100cm^3. The density of the copper is 8900 kgm^-3 and tin 7300kgm^-3. How much volume of each metal must be used if the alloy is to have the density of 7620kgm^-3?

Homework Equations

Volume x density = Mass

The Attempt at a Solution

I've worked out the total mass, (100x10^-6 x 7620 = 0.762kg).

From here I try to re-arrange this equation x + y = 1x10^-4m^3. X = The volume of copper and Y = the volume of tin.

I don't know them, so I get x = 1x10^-4m^3 - y
0.762kg/8900kgm^-3 = 1x10^-4m^3 - y
8.89...x10^-5m^3 = 1x10^-4 - y (Doing the calculation of 0.762kg/8.89...m^3)
-1.101123...x10^-5m^3 = -y Subtracting 1x10^-4 from both sides
1.1011..x10^-5m^3 = y Times by minus 1
Then (1x10^-4) - y = 8.898876x10^-5m^3
x = 8.898876x10^-5m^3

Copper volume(x) = 8.898876x10^-5m^3
Tin volume(y) = 1.1011..x10^-5m^3

I think that is right, but I'm doubting if it is right or not. Reasoning being I divided the total mass by the density of x (copper). Can I do that to work out the volume of x and sub it into that equation? Because logically in my head that does not work. Unless I'm mis-understanding something.

 

[DrClaude]

I think that is right, but I'm doubting if it is right or not.

What is the density of that alloy?

Please use units in your calculations.

 

[lendav_rott]

100 cm³ is how many m³? Please write out your calculations in a friendly manner. It is extremely difficult to follow what exactly is happening.

For the sake of the assignment, you do not need 2 variables. You can express one through the proportion of the other. For example, If A's density was 100 units and B's density was 400 units -> taking x as the density of A, B's density would be 4x. You have to express volumes, but the idea is the same.

It means the proportion of either material's mass in the alloy in relation to the alloy's total mass is exactly the same as the proportion of the volumes of either material in relation to the alloy's total volume respectively. Express the proportion of volumes through what you know. The only formula you need to tackle this assignment is$$\rho=\frac{m}{V}$$which you have. It is just a matter of manipulating the figures to make them give you what you are looking for.

If you think your answer is right, then check it.
Right now, only the volume of Cu weighs more than the required volume of the alloy.

I got 2 * 10^-5 m³ of copper.

 

[dominicwild]

100 cm³ is how many m³? Please write out your calculations in a friendly manner. It is extremely difficult to follow what exactly is happening.

For the sake of the assignment, you do not need 2 variables. You can express one through the proportion of the other. For example, If A's density was 100 units and B's density was 400 units -> taking x as the density of A, B's density would be 4x. You have to express volumes, but the idea is the same.

It means the proportion of either material's mass in the alloy in relation to the alloy's total mass is exactly the same as the proportion of the volumes of either material in relation to the alloy's total volume respectively. Express the proportion of volumes through what you know. The only formula you need to tackle this assignment is$$\rho=\frac{m}{V}$$which you have. It is just a matter of manipulating the figures to make them give you what you are looking for.

If you think your answer is right, then check it.
Right now, only the volume of Cu weighs more than the required volume of the alloy.

I got 2 * 10^-5 m³ of copper.

I've been trying to re-arrange $$\rho=\frac{m}{V}$$ But every time It either gets too messy, or end up getting stuck on trying to get what I want from it. I've been messing around with equations and re-arranging but it's got me no where. Sorry if I'm a a bit slow on understanding this.

Thank you for the help too.

 

[Chestermiller]

These approaches assume that there is no volume change upon mixing the pure constituents, or, equivalently, that the partial mass volumes of the metals in the solution is the same as their pure specific volumes. I'm not sure whether this is a valid assumption. At the very least, it should be stated as an assumption.

Chet

 

[dominicwild]

These approaches assume that there is no volume change upon mixing the pure constituents, or, equivalently, that the partial mass volumes of the metals in the solution is the same as their pure specific volumes. I'm not sure whether this is a valid assumption. At the very least, it should be stated as an assumption.

Chet

Yeah I don't think that the question goes quite to the extent of that depth. Assuming that those are the values without any volume change.

 

[lendav_rott]

Let's start with what you don't know in this assignment. These are the respective volumes of the metals in the alloy.

Let's call the volume of Cu in the alloy X m³.

What is the remaining volume, well, it has to be the Sn in that case, hasn't it? The total volume is given, hence

the volume of Sn is (10^-4 - X)m³.

How would you express how many % of it is Cu in the total volume of the alloy?

 

[dominicwild]

Let's start with what you don't know in this assignment. These are the respective volumes of the metals in the alloy.

Let's call the volume of Cu in the alloy X m³.

What is the remaining volume, well, it has to be the Sn in that case, hasn't it? The total volume is given, hence

the volume of Sn is (10^-4 - X)m³.

How would you express how many % of it is Cu in the total volume of the alloy?

Well to get the percentage you'd (X/10^-4) x 100? I think?

 

[lendav_rott]

Correct, it is Cu's volume in the total volume of the alloy. Now, don't multiply it by 100 to get the percentage, I just thought maybe if I used the word percentage you would click.
What about the mass? Cu's mass divided by the total mass of the alloy is Exactly the same.Let's call Cu's mass in the alloy m_Cu Equate those two to get
$$\frac{X}{V_{alloy}}= \frac{m_{Cu}}{m_{alloy}}$$
There is something you must do still, if I told you what, though, I would have solved the assignment for you.

 

[nasu]

The volume and mass fractions are not the same.
Think that you have equal volumes of aluminum and lead. The volume fraction of lead will be 1/2.
But the mass fraction of lead will be much higher.

What yo can do is to express the density of the alloy in terms of the densities of the components and the volume fraction.
You can start with the definition:
d_alloy=m_alloy/V_alloy

And then m_alloy=m1+m2=d1*V1+d2*V2
and V_alloy=V1+V2.
Then use the volume fraction to express all volumes in term of V_alloy.

 

[lendav_rott]

Sorry, nasu is right - I had deduced that in my own calculations as well, but forgot to mention it. I said they were directly proportionate, but no, they are just proportionate. Similar by 1 factor of k.

In the end all will be the same and you have 2 * 10^-5m³ of Cu and 8 * 10^-5m³ of Sn in the alloy as I said before, turns out my English is a tad sloppy.

 

[Eberhard]

Homework Statement

"An alloy of copper and tin has a volume of 100cm^3. The density of the copper is 8900 kgm^-3 and tin 7300kgm^-3. How much volume of each metal must be used if the alloy is to have the density of 7620kgm^-3?

Homework Equations

Volume x density = Mass

The Attempt at a Solution

I've worked out the total mass, (100x10^-6 x 7620 = 0.762kg).

From here I try to re-arrange this equation x + y = 1x10^-4m^3. X = The volume of copper and Y = the volume of tin.

I don't know them, so I get x = 1x10^-4m^3 - y
0.762kg/8900kgm^-3 = 1x10^-4m^3 - y
8.89...x10^-5m^3 = 1x10^-4 - y (Doing the calculation of 0.762kg/8.89...m^3)
-1.101123...x10^-5m^3 = -y Subtracting 1x10^-4 from both sides
1.1011..x10^-5m^3 = y Times by minus 1
Then (1x10^-4) - y = 8.898876x10^-5m^3
x = 8.898876x10^-5m^3

Copper volume(x) = 8.898876x10^-5m^3
Tin volume(y) = 1.1011..x10^-5m^3

I think that is right, but I'm doubting if it is right or not. Reasoning being I divided the total mass by the density of x (copper). Can I do that to work out the volume of x and sub it into that equation? Because logically in my head that does not work. Unless I'm mis-understanding something.

You can check your result by calculating the density with your values and checking if the correct result comes out.

 

[Eberhard]

I find it clearer if you calculate with symbols and only insert the values in the final formula.

 

[nasu]

@Eberhard Do you realize that this is an 8 years old thread?

 

[jim mcnamara]

Thread locked.