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

Let T be the equilibrium temperature when sphere passes through the ring and their equal diameters be k.

Using coefficients of expansion of Copper and Aluminum

The linear expansion equation is: ΔL=αLiΔT where ΔL = Lf - Li

For Cu

{(k - 2.46)/2.46} = 0.000017 x T

For Al

{(2.46507 - k) / 2.46507} = 0.000023 x T

From above two equation

(2.46507 -2.46)/2.4625 = Tx(0.000040) or

T = 0.00508/(2.5425*0.000040) = 51.47 degree Celsius

Since it's in thermal equilibrium the sume of their heat energy (Q) will equal 0.

Qcu + Qal= 0

Q=mcΔT.

(51.47-0) x 23x0.835 = M*(100-51.47)x0.385

M = (0.835/0.385)*20 g = 52.905 g