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Scott, MIT Graduate
Category: Homework
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Experience:  MIT Graduate (Math, Programming, Science, and Music)
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Themistocle Qin discovers a metal sphere of unusual

Customer Question

Themistocle Qin discovers a metal sphere of unusual properties this Bensalemite sphere has a radius of 80.00 cm and density of 1.753g/cm 3 at 300.00K and atmospheric pressure. Further investigations reveal that Bensalemite transfers heat almost instantaneously, has a large coefficient of thermal expansion (x=3.071 x 10-2/C 0), and is elastic but rather incompressible (bulk modulus 9737GPa). Qin uses cold and pressure to compress the Bensalemite sphere. When the sphere is compressed under a pressure of 2001.0403 atmospheres and cooled to the boiling point of liquid nitrogen (-195.79o C) find the following for the Bensalemite sphere.
A. Volume
B. Density
C. Speed of sound. This is for highschool Physic, thank you, deborah
Submitted: 1 year ago.
Category: Homework
Customer: replied 1 year ago.
Make sure all work is shown. Thanks deborah
Expert:  Ryan replied 1 year ago.

Hi,

Thank you for using the site. I'll be happy to help with this problem.

I'll have the solution (with work shown) posted for you just as soon as I finish typing it up.

Thanks,

Ryan

Expert:  Ryan replied 1 year ago.

Hi again,

Can you please check the numbers that were given in the problem statement?

With the values that are given, the change in volume due to cooling exceeds the original volume of the sphere, which doesn't make any sense.

Thanks,

Ryan

Customer: replied 1 year ago.
Okay I'll check and get back to you...thanks
Expert:  Ryan replied 1 year ago.

OK, thanks.

Customer: replied 1 year ago.
All the figures are the same as on my sheet...do you think the x=3.701 should be a=3.701 ?
Customer: replied 1 year ago.
Do you know what number could be wrong..
Expert:  Ryan replied 1 year ago.

Yes, the coefficient of thermal expansion is usually assigned the variable name "alpha", which is a Greek letter "a".

But it's the values rather than the variable names that are a problem here.

Does this equation look familiar:

∆V = 3V(alpha)∆T

That is the usual formula for the change in volume. Perhaps your course is teaching something different.

The most suspicious number is ***** value of alpha. Most coefficients of thermal expansion are in the range of 1 x 10^-6 to 200 x 10^-6, so a value of 3.071 x 10^-2 is HUGE.

Customer: replied 1 year ago.
i don't have that equation! My iPad doesnt allow me to show the little -2 next to the 10...
Customer: replied 1 year ago.
I don't know what to do...
Customer: replied 1 year ago.
Dr W doesn't want me to use any possible quantum or relativistic effects that have not been studied yet
Expert:  Ryan replied 1 year ago.

That equation is little different from linear thermal expansion. It only has the "3" added to it because it is calculating volume instead of length.

Is there an example problem from your textbook or from class that would help me know what approach your instructor is expecting you to use?

Customer: replied 1 year ago.
You have to find two different volumes (volume initial and volume final).
Customer: replied 1 year ago.
There isn't a problem like it.
Expert:  Ryan replied 1 year ago.

That's the difficulty that I am having.

The initial volume is:

V(initial) = (4/3)*pi*(80 cm)^3 = 2,144,660.58 cm^3

According to the formula (just dealing with the thermal compression for now):

∆V = V(final) - V(initial) = 3 * V(initial) * alpha * (Tf - Ti)

which gives:

V(final) = V(initial) + 3 * V(initial) * alpha * (Tf - Ti)

After I plug in all the values, I get:

V(final) = 2,144,660.58 + 3(2,144,660.58)(3.071 x 10^-2)(-195.79 - 28.65)

V(final) = -42,201,895.70 cm^3

But the final volume can't be negative, so something isn't right.

Customer: replied 1 year ago.
Do you think my teacher is being weird...I don't know what to do.
Expert:  Ryan replied 1 year ago.

I have no idea what your teacher may be trying to do. But I have seen a variety of teachers teach unusual things, so there is always the possibility that I'm just not familiar with the particular approach that you are being taught.

What textbook are you using? Perhaps I can find it online.

Customer: replied 1 year ago.
I'll get to you tomorrow...but he made this problem up.
Expert:  Ryan replied 1 year ago.

Ok. Perhaps it's just a typo somewhere, or perhaps he didn't work the problem out in advance. These things happen.

If you get any new information, please feel free to let me know. I'll be happy to revisit the problem.

Customer: replied 1 year ago.
Thanks Ryan...I'll talk to the teacher tomorrow and let you know.
Expert:  Ryan replied 1 year ago.

You're welcome.

Customer: replied 1 year ago.
Ryan,
The teacher isn't changing anything and I've double proofed my question. Should I submit your answer anyway? What about the other two answers needed?
Expert:  Ryan replied 1 year ago.

The difficulty is that the three answers depend on each other. You need the answer to the volume question to calculate the new density for part 2, and then you need that new density to calculate the speed of sound in part 3.

It would be one thing if we were coming up with a volume that was just unusual, like a 95% change in volume. But the result that we are getting is an impossible result. The volume cannot be negative, so the minimum volume would have to be 0, which would make the density infinite (since you can't divide by 0), and then the speed of sound would be 0, since you would be dividing the bulk modulus by an infinite density.

In essence, this process would be creating a black hole. Perhaps that is what your instructor is leading you toward, even though that seems a bit silly. (Does he seem like the sort that would give you a "trick" question like this?)

I can write up a solution that way if you'd like. Alternatively, if you can send me some of the equations from the section of your textbook where this kind of topic is covered, perhaps I can figure out what they are thinking.

Customer: replied 1 year ago.
Ryan, thanks for trying...sorry it didn't work out.
Thanks,
Deborah
Expert:  Ryan replied 1 year ago.

You're welcome.