Thermodynamics Linear Expansion Problem

AI Thread Summary
The problem involves determining the common temperature at which a brass plug will fit into an iron ring, given their initial diameters and coefficients of linear expansion. The initial diameters are 8.749 cm for the plug and 8.739 cm for the ring, with a required expansion of 0.01 cm. The equations for linear expansion are applied, leading to the equation 8.749(1+19E-6*ΔT) = 8.739(1+12E-6*ΔT). After calculations, the result indicates a temperature change of ΔT = -163.05 °C, leading to a final temperature of 183.05 °C. The discussion confirms the correctness of the approach and calculations.
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Homework Statement



A brass plug is to be placed in a ring made of iron. At room temperature (20°C), the diameter of the plug is 8.749 cm and that of the inside of the ring is 8.739 cm. They must both be brought to what common temperature in order to fit?

It's a linear expansion problem. As a a ring is heated, the diameter of the hole expands.
d-plug = 8.749 cm
d-ring = 8.739 cm
⌂diameter = .01cm
α(brass) = 19E-6
α(iron) = 12E-6
⌂T = 20-T or 293-T (⌂T is constant)
To = 20°C = 293°K (Which one should I use?)

T = ?

Homework Equations


L = Lo*(1+α⌂T)


The Attempt at a Solution


Set the L's of the brass and iron rings equal to each other.
Lo*(1+α⌂T)=Lo*(1+α⌂T)
8.749(1+19E-6*⌂T) = 8.739(1+12E-6*⌂T)
8.749 + .000166⌂T = 8.739 + .000105⌂T
.01 = -.000061⌂T
⌂T = -163.05 °C (I think it's Celsius)
20-(-163.05) = 183.05°C

I attempted a few different methods to solving this, but they proved to be wrong. Am I correct in my thinking?
 
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You work looks good to me.
 
Okay, thank you. I just needed to check that I had the right work and thinking.
 

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