- #1

paulie

- 13

- 0

## Homework Statement

Find the common temperature at which the inner diameter of the ring is 0.05% larger than the diameter of the sphere.

Given:

Lead sphere

d = 5cm

T

_{o}= 70°C

∝ = 29x10

^{-6}/C°

Steel ring

d = 4.9975cm (the sphere is 0.05% larger than the inner diameter of the ring, hence this value)

T

_{o}= 70°C

∝ = 12x10

^{-6}/C°

## Homework Equations

A

_{f}= A

_{o}(1+2∝Δt)

Δt = T

_{f}- T

_{o}

## The Attempt at a Solution

Equate both final area of sphere (S) and ring (R)

A

_{FR}= A

_{FS}(1.0005)

Substitute

A

_{f}= A

_{o}(1+2∝Δt)

A

_{oR}(1+2∝

_{R}Δt) = (1.0005)A

_{oS}(1+2∝

_{S}Δt)

Arrange equation to find Δt

Δt = (1.0005)(A

_{oS}) - (A

_{oR}) / 2(A

_{oR})(∝

_{R}) - (1.0005)(2)(A

_{oS})(∝

_{L})

Enter the values

Δt = (1.0005)(π(5cm/2)

^{2}) - (π(4.9975cm/2)

^{2}) / 2(π(4.9975cm/2)

^{2})(12x10

^{-6}/C°) - 2(1.0005)(π(5cm/2)

^{2})(29x10

^{-6}/C°)

The answer is

Δt = -44.0416 C°

Which is incorrect because the Final temperature is supposed to be 11.2766°C, using my answer and finding for the final temperature with 70°C as initial.

T

_{f}= 25.9584°C

EDIT:

I've found out that if I multiplied 1.0005 on the diameter of lead sphere (5cm) immediately before others, I'll get an answer of Δt = -58.6820 C° where the final temperature will be 11.3180°C which is pretty close but not the exact answer :(