Deals With Linear Expansion. I have no idea where to start. Any Help would be ap

Question

Deals With Linear Expansion. I have no idea where to start. Any Help would be appreciated!

A 16.0 g copper ring at 0.000degreeC has an inner diameter of D = 2.54000 cm. An aluminum sphere at 98.0 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 alpha Al = 23 times 10-6/C degree and alpha Cu = 17 times 10-6/C degree, respectively. . What is the mass of the sphere? kg

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- 113)

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- 113)

Solve for 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.
.016(387)Tf + m(900)(Tf-113)=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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