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)

r2 = r1 + a*(T-To)r1 = 4.42 cm.

4.40 + 1.3*10^-4(T-20)4.40 = 4.42
4.40 + 5.72*10^-4(T-20) = 4.42
5.72*10^-4(T-20) = 4.42-4.40 = 0.02
5.72*10^-4T - 114.4*10^-4 = 0.02
5.72*10^-4T = 0.02 + 0.01144 = 0.03144
T = 314.4*10^-4/5.72*10^-4 = 54.97o

To solve this problem, we can use the concept of area expansion and the formula for linear expansion.

The linear expansion formula is given by:
ΔL = αLΔT

Where:
ΔL is the change in length,
α is the coefficient of linear expansion,
L is the original length,
ΔT is the change in temperature.

In this problem, we need to find the final temperature, so we rearrange the formula to solve for ΔT:
ΔT = ΔL / (αL)

Now, since we are treating this as an area expansion problem, we need to consider the change in area instead of length. The change in area is given by:
ΔA = 2αALΔT

Where:
ΔA is the change in area,
2αA is the linear expansion coefficient for area expansion (2αA is used because the area of a circle is proportional to the square of the radius, and the linear expansion coefficient is proportional to the radius).

We are given:
Initial diameter, D1 = 4.40 cm
Final diameter, D2 = 4.42 cm
Room temperature, T1 = 20ºC

First, let's calculate the initial area, A1:
A1 = π(D1/2)^2

Then, calculate the final area, A2:
A2 = π(D2/2)^2

Next, let's calculate the change in area, ΔA:
ΔA = A2 - A1

Now, we can plug the values into the formula for change in area to find ΔT:
ΔT = ΔA / (2αA)

Finally, to find the final temperature, T2, we add ΔT to the initial temperature, T1:
T2 = T1 + ΔT

By following these steps, you should be able to calculate the final temperature to which the frames must be heated in order to insert lenses of a larger diameter.