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 answer this question, we need to use the concept of thermal expansion and the equation for linear expansion.
The equation for linear expansion is:
ΔL = α * L0 * ΔT
where:
ΔL is the change in length,
α is the coefficient of linear expansion,
L0 is the initial length, and
ΔT is the change in temperature.
In this case, we are given the coefficient of linear expansion (α = 1.30 * 10^(-4) °C^(-1)), the initial radius of the lens holes (2.27 cm), and the new radius of the lenses (2.28 cm).
We want to find out the change in temperature (ΔT) required to expand the frames to accommodate the larger lenses. Since we know the change in length (ΔL), we can use the equation for linear expansion to find ΔT.
First, let's calculate the change in length (ΔL) of the frames based on the change in radius (Δr) of the lens holes:
ΔL = α * L0 * ΔT
Since the lens holes are circular, the change in length is twice the change in radius (ΔL = 2 * Δr). So we can rewrite the equation as:
2 * Δr = α * L0 * ΔT
Solving for ΔT:
ΔT = (2 * Δr) / (α * L0)
Now, we can substitute the given values:
Δr = 2.28 cm - 2.27 cm = 0.01 cm
α = 1.30 * 10^(-4) °C^(-1)
L0 = 2.27 cm
ΔT = (2 * 0.01 cm) / (1.30 * 10^(-4) °C^(-1) * 2.27 cm)
Calculating the change in temperature (ΔT):
ΔT ≈ 245.87 °C
Therefore, the frames must be heated to approximately 245.87 °C in order to accommodate lenses with a radius of 2.28 cm.