A pair of eyeglass frames made of plastic with a coefficient of linear expansion of
1.30x10-4
(Cº
)
-1
have circular lens holes of diameter 4.40 cm at room temp (20º
C).
To what final temperature must the frames be heated in order to insert lenses
4.42 cm in diameter? (Hint: treat this as an area expansion problem)
To solve this problem, we need to calculate the final temperature to which the frames must be heated in order to insert lenses with a diameter of 4.42 cm. Given that the frames are made of plastic with a coefficient of linear expansion of 1.30x10^-4 (Cº)^-1, we can treat this as an area expansion problem.
First, let's calculate the initial area of the lens holes. The diameter of the lens hole is given as 4.40 cm, so the initial radius is half of that, which is 2.20 cm. The initial area of the lens hole can be calculated using the formula for the area of a circle: A = πr^2.
Initial Area = π(2.20 cm)^2
Next, we can find the final radius of the lens hole after it expands to accommodate lenses with a diameter of 4.42 cm. The final radius can be calculated using the formula for linear expansion: ΔL = αLΔT, where ΔL is the change in length, α is the coefficient of linear expansion, L is the initial length, and ΔT is the change in temperature. Since the diameter of the lens hole is changing, we can treat it as the initial length.
ΔL = r_f - r_i
where r_f is the final radius and r_i is the initial radius.
ΔL = (4.42 cm/2) - (2.20 cm)
Now we can rearrange the formula to solve for ΔT:
ΔT = ΔL / (αL)
Since the length in this case is the radius, L is equal to r_i:
ΔT = ΔL / (αr_i)
Finally, we substitute the values into the formula and solve for ΔT:
ΔT = [(4.42 cm/2) - (2.20 cm)] / (1.30x10^-4 (Cº)^-1 * 2.20 cm)
Solving this equation will give us the change in temperature (ΔT) required to expand the lens hole to accommodate lenses with a diameter of 4.42 cm.
I hope this explanation helps you understand how to solve this problem.