# Please help me with my physics homework



## Ewks (Nov 24, 2008)

So I have some physics homework which I can't solve. Here they are

1. When the tempature of a coin rises 98 °C the diameter grows 0.18%
a) how much does the mass grow?

b) what about the area? ( I think I got this solved, but I'm not sure)


2. There's 0.20 ml of etanol in a thermometer in the tempature 0 °C. How many mm does the surface rise in the tube when the tempature rises 10 °C? The meter is made of class and the inner diameter of the tube is 0,55 mm.


Hope you guys can understand my translation. I really tried these but I just wasn't able to do them.


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## nitrocan (Nov 24, 2008)

The mass doesn't grow of course.


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## Ewks (Nov 24, 2008)

nitrocan said:


> The mass doesn't grow of course.



I thought of that but then I thought that what if when the coin expands some little particles would go in to the tiny space left there.


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## nitrocan (Nov 24, 2008)

Ewks said:


> nitrocan said:
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> > The mass doesn't grow of course.
> ...



The MASS of the coin won't change.


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## (X) (Nov 24, 2008)

I agree with nitrocan


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## Ewks (Nov 24, 2008)

Thanks. anyone know anything about the other two?


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## nitrocan (Nov 24, 2008)

I think we're missing some information on the second question (for example: what's the expansion constant of etanol?, or some other info about the thermometer)


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## Ewks (Nov 24, 2008)

I'm not sure what you mean by the "expansion constant" (as my knowledge of english words is limited) but I know that the coefficient of thermal expansion of etanol is 1.10*10^-3. The termometer is just basic meter which uses the thermal expansion and the "tube" in which the etanol is in has a diameter of 0.5mm. and the coefficient of thermal expansion of class is 8*10^-6

I hope that helps...


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## JBCM627 (Nov 24, 2008)

Pretty sure this is right... may not be though.



Spoiler



For part b) of #1, the initial area should be pi*(r)^2, and after being heated, it should be pi*(1.0018*r)^2, so the area increases by a factor of 1.0018^2 = 1.0036

The constant of expansion for glass is small enough that we can probably just neglect it - the ethanol is the important one.
Per wikipedia, the coefficient for ethanol is 750E-6/°K, so ∆V/V = 750E-6*∆T, where V = .2mL and ∆T = 10°C. So ∆V = .0015mL. So .0015mL=.0015cm³=1.5mm³. ∆V=(pi*r^2)*∆h, so 1.5 = pi*(.55/2)^2*∆h... so ∆h=6.31mm.


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## nitrocan (Nov 24, 2008)

Ewks said:


> I'm not sure what you mean by the "expansion constant" (as my knowledge of english words is limited) but I know that the coefficient of thermal expansion of etanol is 1.10*10^-3. The termometer is just basic meter which uses the thermal expansion and the "tube" in which the etanol is in has a diameter of 0.5mm. and the coefficient of thermal expansion of class is 8*10^-6
> 
> I hope that helps...



No I was just trying to tell you what I mean, your terms are correct 

These should be supplied with the question though. After this, you find what the diameter after the heating will be, then calculate the volume of etanol and calculate the height it should be at, then subtract that from the first height.


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## rjohnson_8ball (Nov 24, 2008)

Area increases by the square of the linear size. So an 0.18% increase in diameter means the new area becomes (1.0018)^2 of the original. Subtract 1 from that (and divide by 100) to get the percent increase of 0.36%. This can be estimated easily: (1 + x)^2 = 1 + 2x + x^2 =~ 1 + 2x when x is very small. So if the linear size change is small, the surface area is double the the linear size change.


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## nitrocan (Nov 24, 2008)

If the diameter expands by its 0.0018, then the radius expands by its 0.0018 as well.
If the previous area was π*r^2, then it becomes π*(r*1.0018)^2
π*r^2*(1.0018)^2 - π*r^2 = π*r^2*0.00360324 (nearly 0.36 percent)
so the area increase is ~0.36%


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## Ewks (Nov 25, 2008)

Thanks guys  I got the b) right on my own and the 2. was actually quite easy, stupid me


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## JTW2007 (Dec 1, 2008)

I believe the mass of the coin would change. Of course the amount of matter wouldn't change, but mass is the amount of matter *relative to the volume.*


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## JBCM627 (Dec 1, 2008)

JTW2007 said:


> I believe the mass of the coin would change. Of course the amount of matter wouldn't change, but mass is the amount of matter *relative to the volume.*



That would be the density, and the density would decrease.

Now that I think about it though, the mass may increase slightly due to SR?


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## JTW2007 (Dec 2, 2008)

Oh, right. Sorry, I got the two mixed up.


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## nitrocan (Dec 3, 2008)

JBCM627 said:


> JTW2007 said:
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> > I believe the mass of the coin would change. Of course the amount of matter wouldn't change, but mass is the amount of matter *relative to the volume.*
> ...



I don't think that the mass would change, but what is SR?


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## tikva (Dec 3, 2008)

nitrocan said:


> JBCM627 said:
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> > JTW2007 said:
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Special Relativity... used as a buzzword (or a joke)


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## JBCM627 (Dec 3, 2008)

tikva said:


> nitrocan said:
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> > JBCM627 said:
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Yeah, Special Relativity. So what is "temperature" anyway... just kinetic energy right? And per SR, m = Sqrt(E^2 - (pc)^2)/c^2, Where E is the total energy of the system (including kinetic energy). With p constant (at 0), If E increases, so must m.

Also related, I remember asking Tyson about this a while ago... this is why there is no upper limit to temperature: there is no upper limit to mass.

--
Edit, a couple numbers:
If the coin is 3g (a penny) and is 100% copper (close enough to a penny), which has a specific heat of .385J/(g°C), raising it 98 degrees requires 113.2J, which results in a mass increase of 1.26E-15kg.


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## nitrocan (Dec 4, 2008)

JBCM627 said:


> tikva said:
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> > nitrocan said:
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That formula works when the object has reached near light speed. We can't say that my weight increased because I'm running. That difference is too small to care for.


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## JBCM627 (Dec 4, 2008)

nitrocan said:


> That formula works when the object has reached near light speed. We can't say that my weight increased because I'm running. That difference is too small to care for.



That "formula" works at all speeds. The hotter the object is, the faster the individual particles inside it are moving around. The momentum of the coin may be zero, but not the energy. So yes, the difference at such low scales is pretty much negligable, but still there.

And yeah, when you are running your weight probably decreases more due to sweat evaporating than it increases due to relativity.


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