Thermal Expansion

  • #1
236
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Homework Statement


An iron ring of radius 2.1 m is to be fitted on the rim of a wheel of radius 2.121 m. The coefficient of volume expansion for iron is 3.6 x 10-5 K-1. By how much should the temperature of the ring be increased?
(a) 532 °C
(b) 833 °C
(c) 278 °C
(d) 378 °C

Homework Equations


ΔV/V = γΔθ
where ΔV in the change in volume on expansion, V is the original volume, γ is the coefficient of volume expansion & Δθ is the rise in temperature.
3. The attempt at a solution
Well, obviously, ΔV = 0.021 m. Now, if I use the value of γiron in (°C)-1, that is, 12 x 10-6 (not given in the problem), I get the correct answer of 833 °C. However, if I use the given value of 3.6 x 10-5 K-1, I get the answer as 277.78 K, which comes out to be approximately 4 °C.
Why is there a discrepancy?
How do I convert γiron in K-1 to (°C)-1?
 

Answers and Replies

  • #2
ehild
Homework Helper
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1,909

Homework Statement


An iron ring of radius 2.1 m is to be fitted on the rim of a wheel of radius 2.121 m. The coefficient of volume expansion for iron is 3.6 x 10-5 K-1. By how much should the temperature of the ring be increased?
(a) 532 °C
(b) 833 °C
(c) 278 °C
(d) 378 °C

Homework Equations


ΔV/V = γΔθ
where ΔV in the change in volume on expansion, V is the original volume, γ is the coefficient of volume expansion & Δθ is the rise in temperature.
3. The attempt at a solution
Well, obviously, ΔV ΔL= 0.021 m. Now, if I use the value of γiron in (°C)-1, that is, 12 x 10-6 (not given in the problem), I get the correct answer of 833 °C. However, if I use the given value of 3.6 x 10-5 K-1, I get the answer as 277.78 K, which comes out to be approximately 4 °C.
Why is there a discrepancy?
How do I convert γiron in K-1 to (°C)-1?
The problem gives the volume expansion coefficient. You have to fit the ring onto the rim of the wheel, so the length should be increased. The volume of the ring is irrelevant.
You have to calculate the relative change of length, and you have to use the linear thermal expansion coefficient. How is it related to the volume expansion coefficient?
In the formula for the thermal expansion, you have the change of temperature. The temperature difference is the same both in K and °C.
 
Last edited:
  • #3
236
21
The problem gives the volume expansion coefficient. You have to fit the ring onto the rim of the wheel, so the length should be increased. The volume of the ring is irrelevant.
You have to calculate the relative change of length, and you have to use the linear thermal expansion coefficient. How is it related to the volume expansion coefficient?
Oh, holy cow! Change in circumference, you're damn right! I'm a dumbass.
γ=3α. I get it now.
In the formula for the thermal expansion, you have the change of temperature. The temperature difference is the same both in K and °C.
I missed that too
 

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