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"A copper sphere of radius 2.000 cm is placed over a hole of radius 1.990 cm in an aluminum plate at 20 deg. C. At what common temperature will the sphere pass through the hole?"

My answer:

((∆T(23* 10-6)(0.01990)- .00010 = ∆T(17* 10-6)(0.02)

((∆T(23* 10-6)(0.01990) - .00010 = ∆T(17* 10-6)(0.02)

(∆T x 4.577x10^-7) – 0.00010 = (∆T x 3.4x10^-7)

- 0.00010 = (∆T x 3.4x10^-7) – (∆T x 4.577x10^-7)

-0.00010 = ∆T x -1.177 x 10^-7

∆T = 849.62

Do my equations and calcs look correct? I believed the answer was in the high 700's range....I also dont know how to bring the 20 degrees into it?

HELP!