At what temperature a gold ring of diameter 6.230 cm be heated so that it can be fitted on a wooden bangle of diameter 6.241 cm? Both the diameters have been measured at room temperature (27^oC)

(Given: coefficient of linear thermal expansion of gold ${\alpha }_L=1.4\times {10}^{-5}{\mathrm{K}}^{-1}$)

Complete Solution:

We use the formula for thermal expansion:

ΔL = L₀ α ΔT

Where:

• ΔL = change in length (change in the diameter of the gold ring),
• L₀ = initial diameter of the gold ring = 6.230 cm,
• α = coefficient of linear expansion for gold = 1.42 × 10^-5 °C^-1,
• ΔT = change in temperature (T - T₀).

ΔL = L - L₀ = 6.241 cm - 6.230 cm = 0.011 cm

ΔT = ΔL / (L₀ α)

Substitute the values:

ΔT = 0.011 / (6.230 × 1.42 × 10^-5)

ΔT = 0.011 / 8.8426 × 10^-5 ≈ 124.4 °C

T = T₀ + ΔT = 27 + 124.4 = 151.4 °C

The gold ring must be heated to approximately 152 °C to fit onto the wooden bangle.