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

at equilibrium temperature

diameter of sphere = diameter of ring

d*(1 + alpha_A*(100-t)) = D*(1+ alph_C*(t-0))

alpha_A = coefficeint of expansion of Aluminium = 22.2*10^-6

alpha_C = coefficeint of expansion of copper = 16.6*10^-6

2.46507*(1-(22.2*10^-6*(100-t))) = 2.46*(1+(16.6*10^-6*t))

equilibrium temperature t = 28.98 oC

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heat lost by aluminium = heat gained by copper

Mal*Sal*(100-t) = Mc*Sc*(0-t)

specific heat of aluminium = Sal = 910 J/kg K

specific heat of copper = Sc = 386 J/kg K

Mal*910*(100-28.98) = 0.023*386*28.98

Mal = 0.00398 kg = 3.98 grams