Thermal expansion and coefficient of linear expansion

AI Thread Summary
The discussion focuses on calculating the coefficient of linear expansion for a rod that measures 20.05 cm at 20°C and expands to 20.11 cm at 270°C. The change in length of the rod is 0.06 cm, while the steel ruler expands by 0.055 cm due to thermal expansion. The total change in length is 0.115 cm, which is attributed to both the rod and the ruler. Using the formula for linear expansion, the coefficient of thermal expansion for the rod's material is calculated to be 23 x 10^-6 /°C. This analysis highlights the importance of accounting for the expansion of both the object and the measuring device.
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at 20C a rod is exactly 20.05cm long on a steel ruler. Both the rod and the ruler are placed in an oven at 270C, where the rod now measures 20.11cm on the same ruler. What is the coefficient of linear expansion for the material of which the rod is made?

Homework Equations


\DeltaL=L\alpha\DeltaT
L(i)rod=20.05cm
L(f)rod=20.11
L(i)steel=20.05 ??
\DeltaT=250C
\alphasteel=11 x 10^-6


The Attempt at a Solution


Ls=20.05cm(11x10^-6)(250C)
L=.055cm

I don't know what to do now-
 
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The change in length for the rod is 20.11cm-20.05cm plus
the expansion of the steel ruler at its 20.11cm mark:
∆L = La∆T = (20.11 cm)(11 x 10-6 /C˚)(270˚C-20˚C)-- u should assume it
= 0.055 cm
∆L = (20.11cm-20.05cm) + 0.055 cm = 0.115 cm
The coefficient of thermal expansion of the material the rod
is made of is:
a = ∆L/L∆T=23 x 10-6 /C˚
 
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