A 21.0 g copper ring at 0.000 A 21.0 g copper ring at 0.000 degree C has an inne

Question

A 21.0 g copper ring at 0.000

A 21.0 g copper ring at 0.000 degree C has an inner diameter of D = 2.54000 cm. An aluminum sphere at 108.0 degree C has a diameter of d = 2.54508 cm. The sphere is placed 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 the equilibrium temperature, and the coefficients of linear expansion of aluminum and copper are ? Al = 23 times 10 -6/C degree and ?Cu = 17 times 10 -6/C degree, respectively. . What is the mass of the sphere?

Explanation / Answer

The linear expansion equation is: ?L=?Li?T
So for Copper:
Lf-2.54= (17 10-6)(2.54)(Tf-0)

for Aluminum:
Lf-2.54508= (23 10-6)(2.54508)(Tf- 108)

For the sphere to pass through, Lf has to be the same for both. Using copper we see that
Lf= 2.54+(17 10-6)(2.54)(Tf)
Plug this value into Lf in the Aluminum equation.
2.54+(17 10-6)(2.54)(Tf) -2.54508= (23 10-6)(2.54508)(Tf- 108)

Tf=.

Since it's in thermal equilibrium the sume of their heat energy (Q) will equal 0.
Qcu + Qal= 0
Q=mc?T.
c is the specific heat for the material in J/kgC.
.024(387)Tf + m(900)(Tf-108)=0
Input your value for Tf that you solved and then you can solve for m (the mass of the sphere)

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