A 16.0 g copper ring at 4.8 A 16.0 g copper ring at 4.8 degree C has an inner di

Question

A 16.0 g copper ring at 4.8

A 16.0 g copper ring at 4.8 degree C has an inner diameter of D = 2.78132 cm. A hollow aluminum sphere at 81.0 degree C has a diameter of d = 2.78524 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, in grams? The linear expansion coefficient of aluminum is 23.0 x 10-6 /C degree , the linear expansion coefficient of copper is 17.0 x 10-6 /C degree , the specific heat of aluminum is 900 J/kg·K, and the specific heat of copper is 386 J/kg·K.

Explanation / Answer

copper gets hotter, Al gets colder

M(cu)*c(cu)*(T-4.8) = M(al)*c(al)*(81-T)

At equilibrium their diameters are the same

D +D*alpha(cu)*(T-4.8) =d-d*alpha(al)*(81-T)

D -D*alpha(cu)*4.8 -d +d*alpha(al)*81 = T(d*alpha(al) -D*alpha(cu))

2.78132 -2.78132*17*10^-6*4.8 -2.78524 +2.78524*23*10^-6*81 =T*(2.78524*23*10^-6 - -2.78132*17*10^-6)

1.0419*10^-3 =T*1.6778*10^-5

T =62.1 degree C

Now

M(cu)*c(cu)*(T-4.8) = M(al)*c(al)*(81-T)

c(cu) =0.386 J/g/k

c(al) =0.9 J/g/k

(http://hyperphysics.phy-astr.gsu.edu/hbase/tables/sphtt.html)

16*0.386*(62.1-4.8) =M(al)*0.9*(81-62.1)

353.88 =M(al)*17.01

M(al) =20.8 grams

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