**Answer:**

**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**