An aluminum wing on a passenger jet is 32 m long when its temperature is 25°C. At what temperature would the wing be 9 cm (0.09 m) shorter?

Answer 1

Metals experience deformation with temperature changes, and in this case we are dealing with a linear (one dimensional) deformation due to temperature change. The amount of deformation is dependent on the initial length of the piece being deformed, the magnitude and sign of the change in temperature, and the specific metal's thermal expansion coefficient (a measure of how sensitive the metal is to deformation due to temperature change). The equation for linear deformation is as follows:

∆L=L_oα∆T

L₀ --> the initial length

α --> Thermal Expansion Coefficient (for aluminum, α = 23*10^-6 C^-1)

∆T --> The Change in Temperature

∆L --> The Change in Length (in this case, -0.09 meters)

Solving for ∆T, we find:

∆T = ∆L/(L_oα) = -0.09/(32*23*10^-6)

∆T = -122.3 C

Now, to solve for our final temperature, we just add our change in temperature to our initial temperature

T_f=T₀+∆T=25-122.3=-97.3 C