Question

The Concorde's exterior is made of aluminum, which has acoefficient of linear expansion of At 15 degrees celsius, the Concorde measures 62.1 m in length. When the plane is in flight, friction with the air increases the temperature of the exterior skin to 200 degrees celsius. What is the change in the length of the outer skin of the Concorde?
A) 32 cm
B) 20 cm
C) 24 cm
D) 28 cm
E) 36 cm

Answer 1

Answer: D) 28 cm

Explanation:

This problem can be solved by the following equation:

$\frac{\Delta L}{L_{o}}= \alpha \Delta T$ (1)

Where:

$\Delta L$ is the change in the length of the outer skin of the Concorde

$L_{o}=62.1 m$ is the initial length of the outer skin of the Concorde

$\alpha=2.40(10)^{-5} \°C^{-1}$ is the coefficient of linear thermal expansion for aluminium

$\Delta T=T_{f}-T_{o}=200\°C - 15\°C=185\°C$ is the variation in temperature

Isolating $\Delta L$:

$\Delta L=\alpha \Delta T L_{o}$ (2)

$\Delta L=(2.40(10)^{-5} \°C^{-1})(185\°C)(62.1 m)$ (3)

$\Delta L=0.275 m \frac{100 cm}{1m}=27.5 cm$ (4)

Finally:

$\Delta L=27.5 cm \approx 28 cm$