[SOLVED]Ring and Sphere Linear Expansion

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1. Dec 1, 2014

Haptic9504

1. The problem statement, all variables and given/known data
A 25.0 g copper ring at 0°C has an inner diameter of D = 2.71585 cm. A hollow aluminum sphere at 88.0°C has a diameter of d = 2.72019 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 × 10-6 /C°, the linear expansion coefficient of copper is 17.0 × 10-6 /C°, the specific heat of aluminum is 900 J/kg·K, and the specific heat of copper is 386 J/kg·K.

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c18/fig18_34.gif

2. Relevant equations
$$Q = cm \Delta T$$
$$\Delta L = \alpha \Delta T L_{0}$$

3. The attempt at a solution

Since at equilibrium the sphere passes through the right that would mean that d = D at equilibrium, being that the ring acquires heat and expands while the sphere loses heat and shrinks. I made $$L_{f}$$ be the final diameter of the two.
For the sphere: $$L_{f} - 2.72019 = (2.72019)(23E-6)(T_{f} - 88)$$
For the ring: $$L_{f} - 2.71585 = (2.71585)(17E-6)(T_{f} - 0)$$

Solved for Lf and set them equal, solving for T, getting 68.82 Celcius.
Heat lost by the sphere = Heat gained by ring.
$$cm \Delta T [sphere] = cm \Delta T [ring]$$
$$(900)(m)(88 - 68.82) = (386)(0.025)(68.82)$$
Solving for the mass, m, I get 0.038 kg, which is marked as incorrect.

Not sure what I am doing wrong, be it the whole approach or some minor thing I am overlooking.

2. Dec 1, 2014

Bystander

Your set-up looks good; double check the arithmetic, and if nothing shows up, holler, and I'll double check it.

3. Dec 1, 2014

Haptic9504

I have gone over it a few times, keep getting the same answers. :(

4. Dec 1, 2014

Bystander

I get 71.10 = Tf. You might try looking for places where you or your calculator might have rounded or truncated numbers on intermediate steps.

5. Dec 1, 2014

Haptic9504

Sure enough, that was it. Leave it to my graphic calculator to lead me askew. Thanks!