What is the Work Done by a Heated Metal Rod?

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SUMMARY

The work done by a heated metal rod is calculated using the formula W = 1/2 YAL(α∆t)^2, where W is the work done, Y is the Young's modulus, A is the cross-sectional area, L is the original length, α is the coefficient of linear expansion, and ∆t is the change in temperature. The discussion highlights a discrepancy between the derived formula and a textbook answer, which uses 1/2 YAL(αt)^2, suggesting that the textbook assumes a reference temperature of zero degrees Celsius. The conclusion is that both formulas are valid under different assumptions regarding the reference temperature.

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  • Understanding of Young's modulus (Y) in material science
  • Knowledge of thermal expansion and the coefficient of linear expansion (α)
  • Familiarity with basic physics equations involving work and force
  • Ability to manipulate algebraic expressions for physical formulas
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Homework Statement


a metal rod of length "L",cross sectional area "A",young modulus "Y" and coefficient of linear expansion "α " is heated to "t" degree celsius .The work that can be performed by the rod when heated is

Homework Equations


work done by rod when heated=1/2 force exerted by rod(when heated) multiplied by extension
I know force exerted by rod(when heated)=Y α ∆t multiplied by A
(here ∆t is increase in temperature by due to heating)and extension would be final length -initial length let L' be final length.So extension would be L' - L
As L'=L(1 +α ∆t )
So L' - L=Lα ∆t

The Attempt at a Solution


with the help of equations above we can see
work done by rod when heated=1/2 Y α ∆t A multiplied by Lα ∆t
=1/2 YAL(α ∆t )^2
right?but in my book answer is given 1/2 YAL(αt)^2
which one is right?
 
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It looks like the reference for the exercise writer is zero degrees celsius. You take a more general approach where the environment temperature can be 0 or 20 or whatever.
Apart from that I see no difference between your answer and the book answer ..

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Last edited:
BvU said:
It looks like the reference for the exercise writer is zero degrees celsius.
if that's the case then
BvU said:
no difference between your answer and the book answer ..
but as we can see nothing such is mentioned in the question
 

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