A pair of eyeglass frames is made of epoxy plastic. At room temperature (20.0°C), the frames have circular lens holes 2.50 cm in radius. To what temperature must the frames be heated if lenses 2.53 cm in radius are to be inserted in them? The average coefficient of linear expansion for epoxy is 1.3 ✕ 10-4 (°C) - 1.

T₂ = 114 °C

Explanation:

Area or superficial expansion: can be defined as an increase in area, per unit area per degree rise in temperature. It unit is (1/K) or (1/°C).

β = ΔA / (A₁ΔT) ... equation 1

β = 2α ... equation 2

Area of circle = πr² ... equation 3

Where β = Area expansivity, α = linear expansivity, ΔA = increase in area = (A₂ - A₁), ΔT = change in temperature, A₁ = initial area, A₂ = Final area r = radius,

from the question, The coefficient of linear expansion for epoxy = 1.3 * 10⁻⁴ °C⁻¹,

∴ Coefficient of area expansion for epoxy = 2 * 1.3 * 10⁻⁴ =

β = 2.6 * 10⁻⁴ °C⁻¹,

Using equation 3 to calculate for area, and taking (π = 3.143)

r₁ = 2.5 cm ∴ A₁ = πr₁² = 3.143 * 2.5² = 19.64 cm².

r₂ = 2.53 cm ∴ A₂ = πr₂² = 3.143 * 2.53² = 20.12 cm².

ΔA = A₂ - A₁ = 20.12 - 19.64 = 0.48 cm².

Making ΔT the subject of the relation in equation 1.

ΔT = ΔA / (βA₁) ... equation 4

Substituting the values above into equation 4,

ΔT = 0.48 / (2.6 * 10⁻⁴ * 19.64)

ΔT = 0.48 / (51.064 * 10⁻⁴)

ΔT = (0.48/51.064) * 10000

ΔT = 0.0094 * 10000 = 94 °C

But, ΔT = T₂ - T₁,

Then, T₂ = ΔT + T₁

Where T₁ = 20 °C, ΔT = 94 °C

∴ T₂ = 94 + 20 = 114 °C.

Therefore, temperature at which the frame must be heated if the the lenses 2.53 cm are to be inserted in them = 114 °C