What is the density of pure chromium in g cm^-3?

[Richie Smash]

Homework Statement

This question concerns the resistance wire Nichrome.
''Nichrome is an alloy of nickel (80%) and chromium (20%).
The density of Nichrome is 8.56 g cm^-3.

Diameter/mm ------------------------->1.219 -- 0.914---0.711----0.560------0.457-------0.315
Current carrying capacity/A---------> 15 --------10-------- 7.8------- 4.8------- 3.7---------- 2.0

Determine a likely value for the density of pure chromium in g cm^-3.
Show all steps in your calculation (the density of nickel 8.9 g cm ^ -3).

Homework Equations

D = M/V
volume of a cylinder = pi x Rsquared x height

The Attempt at a Solution

Well, the answer in my book is 7.2 g cm ^ -3.

My book randomly has that table of diameter versus current carrying capacity/A, so my first thought was to substitute the diameter of one of the values, into the volume of a cylinder formula, from this I would have a Volume, which I could substitute into the density formula using the density of Nichrome given, to then find the mass of the Nichrome.

After finding the mass, I would then deduce that since Nichrome is 20% chromium, I would just find 20 percent of the mass, and use the volume I found in my earlier step to find the density of pure chromium.

It looked something like this : R= 1.219/2 = 0.6095mm

Volume of wire= pi x 0.6095^2 x 1 ( I suppose I just assumed the length to be per meter)

= 1.167mm^3

So now to use to density formula, 8.56g cm ^-3 = M/1.167mm^3.
So M = 1.167 x 8.56 = 9.99g...

SO now I found 20% of 9.99... and used it in the density formula again

D= 1.998/1.167 = 1.712g cm ^ -3 was my final answer... but I know I'm horribly wrong.

 

[haruspex]

D= 1.998/1.167

Your 1.998g figure represents the mass of what, exactly? And the volume you divided by is the volume of what?

As recommended to you in another thread, hide your calculator until you have an algebraic answer. Do not even use the numeric values given - create variables for those. Working purely symbolically has many advantages.

Suppose you have a volume V of nichrome. Write expressions for the various masses.

 

[Richie Smash]

Your 1.998g figure represents the mass of what, exactly? And the volume you divided by is the volume of what?

As recommended to you in another thread, hide your calculator until you have an algebraic answer. Do not even use the numeric values given - create variables for those. Working purely symbolically has many advantages.

Suppose you have a volume V of nichrome. Write expressions for the various masses.

Well, I suppose if I have a volume V of Nichrome, the mass would be M=DV, and the other masses would be, m1 and m2, representing the mass of chromium and nickel respective, and I suppose the expressions of those would be (M-m1) and (M-m2).

 

[haruspex]

Well, I suppose if I have a volume V of Nichrome, the mass would be M=DV, and the other masses would be, m1 and m2, representing the mass of chromium and nickel respective, and I suppose the expressions of those would be (M-m1) and (M-m2).

I assume you mean D as the density of the nichrome..
You are given the proportions (by mass) of the two elements. What equations does that lead to?

 

[Richie Smash]

I assume you mean D as the density of the nichrome..
You are given the proportions (by mass) of the two elements. What equations does that lead to?

I'm afraid that I'm not quite sure, I do apologize for my lack of knowledge in this topic.

My guess would be, with D being the density of Nichrome, M being it's mass, and V being it's volume,

M = DV for Nichrome.

Now 20% of Nichrome's mass is Chromium so I suppose in algebraic terms, M1(representing the mass of chromium) = 20/100 x DV = 20DV/100 = DV/5

So therefore the mass of chromium M1 = (DxV)/5.

And for Nickel the mass M2 = (4 x D x V)/5

Am I on the right track here?

 

[Richie Smash]

Does anyone have any ideas? I'm trying to complete a revision course, but I'll move on to other questions for now.

 

[Charles Link]

The information regarding electrical conductivity appears extraneous, and isn't needed to work the problem. When they tell you 20% Chromium and 80% Nickel, they don't specify if it is by mass, by atom, or by volume. Try making the simplifying assumption that it is by volume, and this calculation becomes very straightforward. $\\$ Editing: And yes, that does give the book's answer. $\\$ A hint: Given $D_1,V_1$ and $D_2, V_2$, write out the equation for the density of the resulting mixture when these two things are mixed together. In this case, you can then plug in the numbers, and all you need to do is solve for $D_1$. $\\$ Additional item=when they gave you conductivity information, it looked like they wanted some advanced solid state physics concepts to solve this thing. That, in fact, was simply extra information.

 

[Richie Smash]

Is the equation D3 = (D1 V1+ D2 V2)/V3?

 

[Charles Link]

Is the equation D3 = (D1 V1+ D2 V2)/V3?

Yes. Now is $V1+V2=V3$ ? Also do you know $\frac{V1}{V3}$ and $\frac{V2}{V3}$?

 

[Richie Smash]

Yes I believe that V1 +V2 is V3, and I believe that V1/V2 is 0.2, and V2/V3 is 0.8, which is seeming to me like it's related to the percentages, so my best educated guess that would be a way to verify that the volumes obtained for V1 and V2 are correct?

I'm currently not in school self tutoring for an exam in May, so I don't know everything right now but I think that's a reasonable suggestion.

Edit, I've solved this problem finally, thank you so much Charles, your suggestions help I feel so relieved, I plugged in the values and got exactly 7.2

 

[Charles Link]

Yes I believe that V1 +V2 is V3, and I believe that V1/V2 is 0.2, and V2/V3 is 0.8, which is seeming to me like it's related to the percentages, so my best educated guess that would be a way to verify that the volumes obtained for V1 and V2 are correct?

I'm currently not in school self tutoring for an exam in May, so I don't know everything right now but I think that's a reasonable suggestion.

Edit, I've solved this problem finally, thank you so much Charles, your suggestions help I feel so relieved, I plugged in the values and got exactly 7.2

If Chromium is 20% by volume, that means $\frac{V1}{V3}=.2=\frac{20}{100}=20$% . Similarly, for the nickel, $\frac{V2}{V3}=.8=\frac{80}{100}=80$% . Very good. Glad you got it correct. :) $\\$ Editing: Notice also, if $V3=V1+V2$, then $\frac{V1}{V3}+\frac{V2}{V3}=\frac{V1}{V1+V2}+\frac{V2}{V1+V2}=\frac{V1+V2}{V1+V2}=1=\frac{100}{100}=100$% $\\$ The algebra checks out, just like it should.

 

[haruspex]

the simplifying assumption that it is by volume, and this calculation becomes very straightforward. \\ Editing: And yes, that does give the book's answer.

Yes, that does give the expected answer, but I would regard it as most unusual to specify percentages by volume. You certainly would not want that as the recipe for making a Nichrome since the resulting alloy would depend on the temperature at which you measured the constituents.
In my view, unless stated otherwise, the percentages should imply by mass, and the given answer is wrong.

 

[Charles Link]

Yes, that does give the expected answer, but I would regard it as most unusual to specify percentages by volume. You certainly would not want that as the recipe for making a Nichrome since the resulting alloy would depend on the temperature at which you measured the constituents.
In my view, unless stated otherwise, the percentages should imply by mass, and the given answer is wrong.

@haruspex I think the textbook the OP is using is trying to give the students some simple problems with concepts such as how density is computed. This one also had some rather confusing information, when it introduced information on the conductivity of the wire, but apparently relied on the student knowing that that info should be ignored. In my opinion, this wasn't what I call an ideal problem to teach the student about how density, mass, and volume are related, but apparently we came up what the book was looking for. $\\$ Editing: I just googled the density of Chromium a moment ago, and can you believe, the density is 7.19 grams/cm^3 at 20 degrees Centigrade. The calculations were quite accurate, perhaps with a simplifying assumption.

 

[haruspex]

@haruspex I think the textbook the OP is using is trying to give the students some simple problems with concepts like density. This one also had some rather confusing information, when it introduced information on the conductivity of the wire, but apparently relied on the student knowing that that info should be ignored. In my opinion, this wasn't what I call an ideal problem to teach the student about how density, mass, and volume are related, but apparently we came up what the book was looking for.

Ok, but that is a curious combination: on the one hand so sophisticated as to test that the student understands to ignore extraneous information, but on the other so ignorant as to use the wrong way to specify constituent percentages. In particular, I would not want the student left with the mistaken impression that percentages by volume is standard.

 

[Richie Smash]

To clear up the confusion, the subsequent questions after that one then ask on the relationship of the thickness of the wire and current carrying capacity, based on the information in that table. It was however placed in an odd place, above the question I posted here, so I included the table in the post just to ensure that it was there in case it was needed for this question as well.

The textbook is called ''Heinemann, Physics for CXC''

CXC stands for the Caribbean Examinations Council and the exam I'm writing is in May and it is known as CSEC or Caribbean Secondary Education Certificate,
which is equivalent to 'O' or 'Ordinary level' examinations in the U.K.
The exam will consist of 3 papers. Paper 1 being multiple choice, Paper 2 being Structured Questions, and Paper 3 having structured questions as well but more Lab based questions, for instance Labs such as ''Period of a Pendulum'' or ''Finding the refractive index of Glass.''

The topics covered are Units and Measurements, Forces and Motion, Thermal Physics, Waves and Optics, Electricity and Magnetism, and Physics of the Atom.

So currently I've done all the revision questions for the first four sections, and I'm currently on Electricity and Magnetism, so you'll be seeing me here almost every day, as I only have a short time left.

 

[Charles Link]

Ok, but that is a curious combination: on the one hand so sophisticated as to test that the student understands to ignore extraneous information, but on the other so ignorant as to use the wrong way to specify constituent percentages. In particular, I would not want the student left with the mistaken impression that percentages by volume is standard.

Please read my edited additions to my previous post. Also, it appears the author made up his own numbers here, because a google of NiChrome wire came up as 8.40 gm/cm^3 rather than 8.56 gm/cm^3. I agree with your inputs, but the OP here is studying what is intended to be introductory material. They might have done better to use solutions with different densities or concentrations.

 

[Charles Link]

Let me then give an alternative solution, if I may, that would be the preferred way of computing this quantity: $D3= \frac{D1 \, V1+D2 \, V2}{V1+V2}$. In addition $M2=4 \, M1$ so that $D2 \, V2= 4 \, D1 \, V1$.Numerator and denominator of the first equation can be divided by $V2$, and this can be readily solved for $D1$, (by also using the second equation). $\\$ It is difficult sometimes to assess exactly what level the student is at, but this second solution would be a much improved approach, as @haruspex is correct that these are normally specified 20% vs. 80% by mass.

 

[haruspex]

I just googled the density of Chromium a moment ago, and can you believe, the density is 7.19 grams/cm^3

As I understand it, there is not one standard Nichrome. The are various formulations, some including iron or other metals.

 

[Charles Link]

Suggestion for the OP @Richie Smash : See if you can follow what I did in post 17, and see if you can come up with the answer for $D1$ that you get from those two equations. $\\$ (Do not use $V1/(V1+V2)=.2$ and $V2/(V1+V2) =.8$, because we have determined that that is not what a 20%- 80% composition refers to). $\\$ If my arithmetic is correct, I get a slightly different answer than 7.2 gm/cm^3. I'd be interested in seeing if you get the same thing I did.

 

[Richie Smash]

Suggestion for the OP @Richie Smash : See if you can follow what I did in post 17, and see if you can come up with the answer for $D1$ that you get from those two equations. $\\$ (Do not use $V1/(V1+V2)=.2$ and $V2/(V1+V2) =.8$, because we have determined that that is not what a 20%- 80% composition refers to). $\\$ If my arithmetic is correct, I get a slightly different answer than 7.2 gm/cm^3. I'd be interested in seeing if you get the same thing I did.

Just trying to simplify the expressions first, then I'll plug in the numbers, I seem to be missing something however.

 

[PAllen]

Let me then give an alternative solution, if I may, that would be the preferred way of computing this quantity: $D3= \frac{D1 \, V1+D2 \, V2}{V1+V2}$. In addition $M2=4 \, M1$ so that $D2 \, V2= 4 \, D1 \, V1$.Numerator and denominator of the first equation can be divided by $V2$, and this can be readily solved for $D1$, (by also using the second equation). $\\$ It is difficult sometimes to assess exactly what level the student is at, but this second solution would be a much improved approach, as @haruspex is correct that these are normally specified 20% vs. 80% by mass.

Of course, this then no longer produces the book answer. The nickel will be only 76.9... percent by volume if it is 80 by mass. Of course, often the volume of result is not sum of volumes of constituents, so the result density cannot be computed from constituent densities and percentages without more information.

I found it simpler to set up the by mass case assuming v1 + v2 = 1, without loss of generality. Then you also have

D2 v2 = .8 d3
D1 v1 = .2 d3

 

[Charles Link]

Just trying to simplify the expressions first, then I'll plug in the numbers, I seem to be missing something however.

You have two equations and two unknowns: $D1$ and $r=\frac{V1}{V2}$. It may look like 2 equations and 3 unknowns: $D1$, $V1$ and $V2$, but the ratio of $V1$ to $V2$ shows up in both equations, so it simplifies. It takes a little more algebra to solve it this way, than with the previous assumption of 20-80 by volume, but the previous assumption is really a false assumption that lacks validity. The 20-80 by mass is really the way this problem should be worked.

 

[PAllen]

Here is another formulation of the difference between the two mixing models. Given constituents with density D1 and D2, with fractions f1 and f2 by volume, then total volume may simply be taken as 1 and the fractions used as volumes leading simply to:

f1 D1 + f2 D2 = D3

With f1 and f2 taken by mass, total mass may be taken as 1, with f1 and f2 becoming constituent masses. Then, noting mass/density gives volume, you simply have:

1 / ( (f1/D1) + (f2/D2) ) = D3

Solution of either for any unknown is now trivial.

 

[Charles Link]

Here is another formulation of the difference between the two mixing models. Given constituents with density D1 and D2, with fractions f1 and f2 by volume, then total volume may simply be taken as 1 and the fractions used as volumes leading simply to:

f1 D1 + f2 D2 = D3

With f1 and f2 taken by mass, total mass may be taken as 1, with f1 and f2 becoming constituent masses. Then, noting mass/density gives volume, you simply have:

1 / ( (f1/D1) + (f2/D2) ) = D3

Solution of either for any unknown is now trivial.

@PAllen Both of these formulations are useful. The OP will need a little practice at working through the complete algebra before using more advanced methods that lead a little more quickly to the answer.

 

[Richie Smash]

Here is what I'm doing:

D3 = (D1V1 +D2V2)/(V1+V2)
D3 = (D1V1 +4 D1 V1)/(V1+V2)
D3 = (D1+4D1V1)/V2
D3 = D1
Or I was trying this:
D3 = (D1V1)+D2V2)/V3
D3V3 =D1V1+D2V2
(D3V3)/D2V2= D1V1
{D3(V1+V2)}/D2V2 = D1V1
(D3V1)/D2 = D1 V1
Divide both sides by V1
And I'll get D3/D2 = D1

But that doesn't seem correct either.
Something seems off there you see.

Edit, I am in the introductory stage so I'm now trying hard to figure out your method Charles but I Just don't know where to start, I think what I need is slow small steps.

 

[Charles Link]

Your 3rd line is incorrect. Let me show you what you should get when you divide numerator and denoninator by $V2$: $D3=\frac{5 D1 \,( \frac{V1}{V2})}{(\frac{V1}{V2})+1}$. Also hang on to the second equation $D2 \, V2=4 \, D1 \, V1$, and write it as $r=(\frac{V1}{V2})=\frac{D2}{4 \, D1}$ and substitute it into the equation that I just wrote out for you. $\\$ You got to work at getting the algebra correct in these problems. Much of the introductory physics is algebra.

 

[PAllen]

Consider also my post #21, which is equivalent to what @Charles Link is suggesting, but gives you 3 simple equations in 3 unknowns, which you might find easier to use.

 

[Richie Smash]

Ok Guys I'm getting somewhere, Pallen, I will try your method right after I figure out Charles', and I won't give up until I understand this fully, I'm feeling that this is a higher level that I haven't encountered, but I can do it, I hope you two gentlemen have the patience to bear with me here.

I actually understand how you got r = D2/4D1

Because in the equation D2V2=4D1V1

V1 = (D2V2)/4D1
The you say represent it as V1/V2
V2 would be : (4D1V1)/D2

Now if r = V1V2 then using the two expressions I determined for V1 and V2, r = D2/4D1

Now Here's where I go wrong, I divide both sides by (4D1V1)/ D2 = V2

And the answer i keep getting is D1 = D3/D2

Now that I know how you got r, I need to know why you're using it, it is the ratio of Volume 1 to Volume 2, but I thought we were using mass, this is where I get confused.
If D2V2 = 4D1V1, and both of those are the masses, then wouldn't r be M1/M2?
Because I see it as, the ratio of mass 1 to mass 2, but since you have written it as V1/V2, how does the volume come into play here in relationship to the equation D2V2 = 4D1V1?

 

[Charles Link]

Ok Guys I'm getting somewhere, Pallen, I will try your method right after I figure out Charles', and I won't give up until I understand this fully, I'm feeling that this is a higher level that I haven't encountered, but I can do it, I hope you two gentlemen have the patience to bear with me here.

I actually understand how you got r = D2/4D1

Because in the equation D2V2=4D1V1

V1 = (D2V2)/4D1
The you say represent it as V1/V2
V2 would be : (4D1V1)/D2

Now if r = V1V2 then using the two expressions I determined for V1 and V2, r = D2/4D1

Now Here's where I go wrong, I divide both sides by (4D1V1)/ D2 = V2

And the answer i keep getting is D1 = D3/D2

Now that I know how you got r, I need to know why you're using it, it is the ratio of Volume 1 to Volume 2, but I thought we were using mass, this is where I get confused.

Read post 26 again please. I may have edited it since you last looked at it. Simply substitute $r=\frac{V1}{V2} =\frac{D2}{4 \, D1}$ back into the first equation: $D3=\frac{5 \, D1 \, r}{(r+1)}$. The result that you get will be a somewhat clumsy expression which can then be simplified somewhat, and solved for $D1$. And the algebra in this one is non-trivial as algebra goes, but you need to stick with it, if you want to get the correct answer.

 

[Charles Link]

A note to the OP: I do think your background in algebra might be quite deficient. It is likely you are missing some of the simpler things of solving these expressions. Writing them out in Latex is somewhat clumsy, but using pencil and paper, the above expressions have not been very advanced by any means. Good algebra skills are essential for the coursework that you mentioned in post 15. Good trigonometry and some calculus are also very helpful for some of these topics. You can work to those goals mentioned in post 15, but it's going to require lots and lots of effort=the month of May (your exam) is going to come around very quickly by the time frame I see. There really isn't a simple way to learn some of this material. It comes with hours upon hours of effort and practice.

 

[Richie Smash]

Yes I agree... the exam I'm writing isn't actually a high level, and the book even has harder questions than the exam itself, I agree I'm not understanding some simple algebraic things here... I apologize.

What I understand is up to the second line in my attempt in post 25, and how you got r = D2/4D1 = V1/V2 in terms of the equation D2V2 = 4 D1 V1.
I don't understand how substituting r into the second line of my post 25, i,e D3 = (D1V1+4D1V1)/V1+V2 turns into 5 D1 r/ r+1.

I'm writing the general maths exam as well, the algebra in it is not too complicated, so I don't have experience in these things, but thank you for helping me out Charles and taking the time to guide me, I am going to put in hours and hours in the coming months, but right now I need to spend a long time understanding what you're trying to do, my problem is understanding the simple steps, like ''why does doing this get rid of these terms?'' or '''why is this the ratio of this'' etc.

 

[Charles Link]

The first thing you need is to take the equation that is of the form $D3=\frac{A}{B}$ and divide numerator and denominator by $V2$: $D3=\frac{(\frac{A}{V2})}{(\frac{B}{V2})}$. Then process each part $\frac{A}{V2}$ and $\frac{B}{V2}$ separately. $\\$ You should also have recognized that $D1 \, V1 +4 \, D1 \, V1 =5 \, D1 \, V1$. Doing the complete algebra: $(1) (D1 \, V1)+ (4) (D1 \, V1)=(4+1)(D1 \, V1)=5 \, D1 \, V1$ . $\\$ For the above $A=5 \, D1 V1$, and $B=V1+V2$. It should be a simple matter to divide $A$ by $V2$, and $B$ by $V2$, and then substitute in for $r=\frac{V1}{V2} =\frac{D2 }{4 \, D1}$.

 

[Richie Smash]

Ok I understand how you got A = 5D1V1 and B = V1+V2 {V2 = (4D1V1)/D2
B divided by V2 = V1
and A divided by V2 = 5D1V1 x (D2)/(4D1V1)
= 5D2/4
So the you would have (5D2/4)/V1 {V1 = (D2V2)/4D1}

Now that works out to (5D2)/4 x (4D1)/(D2V2)
= (20D1)/4V2
= 5D1/V2.

Am I getting somewhere?

 

[Charles Link]

Ok I understand how you got A = 5D1V1 and B = V1+V2 {V2 = (4D1V1)/D2
B divided by V2 = V1
and A divided by V2 = 5D1V1 x (D2)/(4D1V1)
= 5D2/4
So the you would have (5D2/4)/V1 {V1 = (D2V2)/4D1}

Now that works out to (5D2)/4 x (4D1)/(D2V2)
= (20D1)/4V2
= 5D1/V2.

Am I getting somewhere?

Progress is somewhat slow. $B$ divided by $V2$ is $\frac{B}{V2}=1+\frac{V1}{V2}$ and then you need to substitute in for the expression for $\frac{V1}{V2}$.

 

[Richie Smash]

Sorry for taking up so much of your time.
I thought that since B=V1+V2 then dividing by V2 would get rid of the two terms and leave V1.
I'm not quite sure how you arrived at your answer, but I assume I solved A wrong as well?