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 0.000 degree C 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

First, find the temperature (Tf) where D = D0 (D = final diamter, D0 = original diameter)

D = D0 (1 + (coefficients of linear expansion of Cu) (Tf - 0 C)

df = di (1 + (coefficients of linear expansion of Al) (Tf - 98 C) ==>

2.54000 (1 + (coefficients of linear expansion of Cu) (Tf - 0 C) = 2.54508 (1 + (coefficients of linear expansion of Al) (Tf - 98 C)

Solve for Tf in the above expression.

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Heat into ring: Q = (specific heat of Cu) (mr)(Tf) where mr = mass of ring

Heat from sphere: lQl = (specific heat of Al)(ms)(Tf - Ti) where ms = mass or sphere...Tf and TI is final and inital Temp.

I will make this easier for you and give you the final expression that will give you the answer:

ms = [(specific heat of Cu)(mr) Tf ] / [(specific heat of Al) (Tf - Ti)] ....remember if you need to, divide grams by 1000 to get kg.

Hope this helps, I would do the work out for you but I do not have a calculator available to me. Please rate!

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