In a class project a student has decided to make a new type of thermometer. The thermometer assembly consists of two plates, one made of aluminum and the other made of steel. Both plates are the same length, b, at room temperature.

When the temperature changes, the plates expand different, causing a pointer to rotate. Each increment on the scale corresponds to a temperature change of 1^o F. The distance, c, on the scale is 2 mm which is very small when compared to the length, L, of the indicator, (c << L).The length of each plate, b, is 50 mm. What must be the length of the indicator? The thermal expansion of the indicator can be neglected. The thermal expansion of aluminum, α_a, is 13×10^-6/^oF and that of steel, α_s, is 6.5×10^-6/^oF.

Change of Position of the Indicator
with Temperature Change

Since each plate is made of different materials their overall length change due to temperature change of 1^oF is different.

     δ_a = b α_a ΔT
         = (50) (13×10^-6) (1)
         = 6.5×10^-4 mm
     δ_s = b α_s ΔT
         = (50) (6.5×10^-6) (1)
         = 3.25×10^-4 mm

Recall, the distance between two consecutive pointer is very small compared to the length of the indicator. So, the length, c, can be assumed to be a straight line. Using similar triangles gives,

     (c - δ_s) / L = (δ_a - δ_s) / a
     (2 - 3.25×10^-4) / L = (6.5×10^-4 - 3.25×10^-4) / 5
     1.9997 / L = 3.25×10^-4 / 5
     L = 5 (1.9997) / ( 3.25×10^-4)
     L = 30,760 mm
     L = 30.76 m