A 30.0 g copper ring at 0°C has an inner diameter of D = 3.56110 cm. A hollow al

Question

A 30.0 g copper ring at 0°C has an inner diameter of D = 3.56110 cm. A hollow aluminum sphere at 97.0°C has a diameter of d = 3.56829 cm. The sphere is placed on top of the ring (see the figure), 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. What is the mass of the sphere? The linear expansion coefficient of aluminum is 23.0 × 10-6 /C°, the linear expansion coefficient of copper is 17.0 × 10-6 /C°, the specific heat of aluminum is 900 J/kg·K, and the specific heat of copper is 386 J/kg·K.

Explanation / Answer

Let T be the temperature reached by both when sphere passes through the ring. Let their equal diameters be d.

Using coefficients of expansion of Copper and Aluminum, we have

{(d - 3.56)/3.56} = 0.000017*T --- (1)

{(3.56829 - d)/3.56829 } = 0.000023*T ---(2)

Adding (1) and (2) give us

(3.56829 - 3.56110)/3.56829 = T*(0.000040) or

T =(3.56829 - 3.56110)/(3.56829*0.000040) = 50.37 deg C

Heat gained = heat lost or

(50.37 - 0)*30*900 = M*(100-50.37)*386 or

M = (50.37 - 0)*30*900 / [(100-50.37)*386]

= 71 g

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