What is the Temperature at which the Gap Between Two Bars will be Closed?

AI Thread Summary
To determine the temperature at which a gap of 1.3x10^-3 m between a brass bar and an aluminum bar will close, the coefficients of linear expansion for both materials are essential. The equation ΔL = αLoΔT is used, where ΔL represents the change in length, α is the coefficient of linear expansion, Lo is the original length, and ΔT is the change in temperature. The total expansion of both bars must equal the gap, leading to the equation 1.3x10^-3 = (19x10^-6 * 2.0 + 23x10^-6 * 1.0)ΔT. Solving this gives a temperature increase of 21.3°C, resulting in a final temperature of 49.3°C when the bars will touch. This method effectively addresses the problem of thermal expansion in different materials.
BOAS
Messages
546
Reaction score
19

Homework Statement



A brass bar and an aluminium bar are each attached opposite each other to immovable walls. (There is a diagram, but I think the idea is clear enough). There is a gap between the two bars of 1.3x10-3m at 28°C.

At what temperature will the gap be closed?

coefficient of linear expansion of brass = 19x10-6 (C°)-1
aluminium = 23x10-6 (C°)-1

Length of brass bar = 2.0m
Length of aluminium bar = 1.0m

Homework Equations



ΔL = \alphaLoΔT

The Attempt at a Solution



I'm not sure where to begin really. Obviously part of the problem is that the two rods are not going to meet in the middle as they expand.

I have 2 unknowns for both rods in the equation for linear expansion, so it's not really helpful to rearrange for ΔT and set them equal to each other.

If you could get me started that would be great,

thanks!
 
Physics news on Phys.org
What must the sum of the ΔL's equal?
 
1.3 x 10^-3 m

Can I say:

ΔL + ΔL = \alphaLoΔT + \alphaLoΔT

1.3x10^-3 = (3.8x10^-5 + 23x10^-6)ΔT

ΔT = 21.3 degrees C

28 + 21.3 = Temp when bars will touch
 
Last edited:
BOAS said:
1.3 x 10^-3 m

Can I say:

ΔL + ΔL = \alphaLoΔT + \alphaLoΔT

1.3x10^-3 = (3.8x10^-5 + 23x10^-6)ΔT

ΔT = 21.3 degrees C

28 + 21.3 = Temp when bars will touch
I didn't check your arithmetic, but that is definitely the way to solve it.
 
  • Like
Likes 1 person
I'll check and double check.

Thanks a lot!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...

Similar threads

Back
Top