A man's gold wedding ring has an inner diameter of 2.0cm at 30∘C.


Part A
If the ring is dropped into boiling water, what will be the change in the inner diameter of the ring?

ΔL =L αΔT =2 •10⁻²•14•10⁻⁶•70=1.96•10⁻⁵m=1.96•10⁻² mm

To determine the change in the inner diameter of the ring when it is dropped into boiling water, we need to consider the property of thermal expansion. When a material is heated, it expands, and when it is cooled, it contracts. The change in length or diameter due to temperature change can be calculated using the following formula:

ΔL = α * L₀ * ΔT

where:
ΔL is the change in length or diameter,
α is the coefficient of linear expansion of the material,
L₀ is the initial length or diameter, and
ΔT is the change in temperature.

For this problem, we know the initial inner diameter of the ring (L₀) is 2.0 cm at 30°C. We need to find the change in the inner diameter (ΔL) when the ring is dropped into boiling water, so we need to know its coefficient of linear expansion (α).

Different materials have different coefficients of linear expansion. For gold, the coefficient of linear expansion is approximately 14.0 x 10^(-6) °C^(-1). We can now plug the values into the formula:

ΔL = (14.0 x 10^(-6) °C^(-1)) * (2.0 cm) * (100 °C - 30 °C)

Calculating this equation:

ΔL = (14.0 x 10^(-6) °C^(-1)) * (2.0 cm) * (70 °C)

ΔL = 1.96 x 10^(-4) cm

Therefore, the change in the inner diameter of the ring when it is dropped into boiling water is approximately 1.96 x 10^(-4) cm.