Question

A 23.0 g copper ring at 0.000°C has an inner diameter of D = 2.46000 cm. An aluminum sphere at 100.0°C has a diameter of d = 2.46507 cm. The sphere is put 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 equilibrium temperature. What is the mass of the sphere?

Answer 1

We know that,

D = D0 [1 + alpha(Cu)(Tf - 0)]

df = di [ 1 + alpha(Al)(Tf- 100) ]

D = df

2.46000 [1 + 17 x 10^-6(Tf - 0)] = 2.46507 [ 1 + 24 x 10^-6(Tf- 100) ]

2.46000 + 0.00004182 Tf = 2.46507 + 0.00005916168 Tf - 0.005916168

0.00001734168 Tf = 2.46 - 2.459153832

Tf = 48.79 deg C

Heat into the ring

Q = m(ring) C(cu) Tf

that from shpere,

Q = m(sph) C(Al) (Tf - Ti)

Ms = C(Cu) m(ring) Tf / C(Al) (Tf - Ti)

Ms = 23 x 0.385 x 48.79 / 0.9 x (100 - 48.79)

Ms = 9.73 g