A pair of eyeglass frames are made of an epoxy plastic (coefficient of linear expansion = 1.30 10-4°C−1). At room temperature (20.0°C), the frames have circular lens holes 2.27 cm in radius. To what temperature must the frames be heated if lenses 2.28 cm in radius are to be inserted into them?

To find the temperature to which the frames must be heated, we need to use the concept of linear expansion. The first step is to determine the change in size of the lens holes when the radius increases from 2.27 cm to 2.28 cm.

Let's denote:
Initial radius (r₁) = 2.27 cm
Final radius (r₂) = 2.28 cm
Coefficient of linear expansion (α) = 1.30 × 10⁻⁴ °C⁻¹

The change in size (ΔL) of the lens holes can be calculated using the formula:

ΔL = α * L * ΔT

where L is the initial length or size, ΔT is the change in temperature, and α is the coefficient of linear expansion.

Since the radius of a circle is a length, we can consider the circumference (C) of the lens holes, which is given by:

C = 2πr

The change in circumference (ΔC) is then:

ΔC = 2π (r₂ - r₁)

Using the relation ΔL = ΔC, we can write:

α * L * ΔT = 2π (r₂ - r₁)

Rearranging the formula to solve for ΔT:

ΔT = (2π * (r₂ - r₁)) / (α * L)

Now we can substitute the given values and calculate ΔT:

L = C = 2πr₁ = 2π * 2.27 cm
r₂ = 2.28 cm
α = 1.30 × 10⁻⁴ °C⁻¹

ΔT = (2π * (2.28 - 2.27)) / (1.30 × 10⁻⁴ * 2π * 2.27)

Simplifying:

ΔT = (2π * 0.01) / (1.30 × 10⁻⁴ * 2π * 2.27)

Cancelling out common terms:

ΔT = 0.01 / (1.30 × 10⁻⁴ * 2.27)

Calculating:

ΔT ≈ 20.31 °C

Therefore, the frames must be heated by approximately 20.31 °C for the lenses with a radius of 2.28 cm to fit into the frame.

Note: It is important to check whether the material can withstand the desired temperature without being damaged.