Question

A 20.0 g copper ring at 0.000°C has an inner diameter of D = 2.54000 cm. An aluminum sphere at 108.0°C has a diameter of d = 2.54508 cm. The sphere is placed on top of the ring, and the two are allowed to come to thermal equilibrium, with no heat lost to the surroundings. The sphere just passes through the ring at the equilibrium temperature, and the coefficients of linear expansion of aluminum and copper are αAl = 23*10^-6/C° and αCu = 17*10^-6/C°, respectively. . What is the mass of the sphere?

Answer 1

The mass of the sphere can be calculated using the equation m = (4/3)πr^3ρ, where r is the radius of the sphere and ρ is the density of aluminum.

The radius of the sphere can be calculated using the equation r = (d - D)/2, where d is the diameter of the sphere and D is the diameter of the ring.

The density of aluminum can be calculated using the equation ρ = m/V, where m is the mass of the aluminum and V is the volume of the aluminum.

The volume of the aluminum can be calculated using the equation V = (4/3)πr^3, where r is the radius of the sphere.

Substituting the values into the equations, we get:

m = (4/3)π((2.54508 cm - 2.54000 cm)/2)^3 * (20.0 g / ((4/3)π((2.54508 cm - 2.54000 cm)/2)^3))

m = 20.0 g * (2.54508 cm - 2.54000 cm)^3 / ((4/3)π(2.54000 cm)^3)

m = 0.0014 g

Therefore, the mass of the sphere is 0.0014 g.