- #1
Haptic9504
- 10
- 0
Homework Statement
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
Homework Equations
Q = cm \Delta T
\Delta L = \alpha \Delta T L_{0}
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 celsius.
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.