A 28.0 g copper ring at 0^ degree C has an inner diameter of D= 2.73426 cm. A ho

Question

A 28.0 g copper ring at 0^ degree C has an inner diameter of D= 2.73426 cm. A hollow aluminum sphere at 87.0^ degree C has a diameter of d = 2.73853 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 times 10^-6/ C degree, the linear expansion coefficient of copper is 17.0 times 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

The linear expansion equation is: L=LiT
So for Copper:
Lf-2.73426= (17*10^-6)(2.73426)(Tf-0)

for Aluminum:
Lf-2.73853= (23*10^-6)(2.73853)(Tf- 87)

For the sphere to pass through, Lf has to be the same for both. Using copper we see that
Lf= 2.73426 +(17*10^-6)(2.73426)(Tf)
Plug this value into Lf in the Aluminum equation.
2.73426 +(17*10^-6)(2.73426)(Tf) -2.73853= (23*10^-6)(2.73853)(Tf- 87)

Solve for Tf.

Tf = 73.3044

Since it's in thermal equilibrium the sume of their heat energy (Q) will equal 0.
Qcu + Qal= 0
Q=mcT.
c is the specific heat for the material in J/kgC.
.028(386)(73.3044) + m(900)(73.3044-87)=0
Input your value for Tf that you solved and then you can solve for m (the mass of the sphere)

So,m = 0.0642 kg

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