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Chapter 13: Problem 17

A steel sphere with radius \(1.0010 \mathrm{cm}\) at \(22.0^{\circ} \mathrm{C}\) must slip through a brass ring that has an internal radius of $1.0000 \mathrm{cm}$ at the same temperature. To what temperature must the brass ring be heated so that the sphere, still at \(22.0^{\circ} \mathrm{C},\) can just slip through?

Short Answer

Expert verified
Answer: The brass ring must be heated to approximately 174.17°C.

Step by step solution

01

Identify the given information and coefficients of linear expansion

We are given the following information: - Steel sphere radius at 22.0°C: \(r_s = 1.0010\,\mathrm{cm}\) - Brass ring radius at 22.0°C: \(r_b = 1.0000\,\mathrm{cm}\) - Initial temperature (\(T_i\)): \(22.0^{\circ}\,\mathrm{C}\) - Coefficient of linear expansion for brass: \(\alpha_b = 19 \times 10^{-6}\,\mathrm{C^{-1}}\) - Coefficient of linear expansion for steel: \(\alpha_s = 11 \times 10^{-6}\,\mathrm{C^{-1}}\) We need to find the final temperature (\(T_f\)) at which the brass ring must be heated so that the steel sphere will just slip through.
02

Calculate the expanded radius of the steel sphere

First, we need to find the expanded radius of the steel sphere at the final temperature. We can use the formula for linear expansion: \(\Delta r_s = r_s \times \alpha_s \times \Delta T\) But we need the sphere to remain at the initial temperature, so its radius won't change: \(\Delta r_s = 0\) So, the steel sphere radius remains constant: \(r_s = 1.0010\,\mathrm{cm}\).
03

Calculate the expanded radius of the brass ring

Next, we need to find the expanded radius of the brass ring when heated. We can use the formula for linear expansion: \(\Delta r_b = r_b \times \alpha_b \times \Delta T\) At the final temperature, the radius of the brass ring should be equal to the radius of the steel sphere for the sphere to just slip through: \(r_s = r_b + \Delta r_b\) Substitute the linear expansion formula into the equation: \(r_s = r_b + r_b \times \alpha_b \times \Delta T\)
04

Calculate the final temperature

Now, we need to solve for the final temperature \(T_f\). First, factor out the final radius of the brass ring: \(r_s = r_b (1 + \alpha_b \times \Delta T)\) Next, solve for \(\Delta T\): \(\Delta T = \frac{r_s / r_b - 1}{\alpha_b}\) And finally, add the initial temperature: \(T_f = T_i + \Delta T\) Plug in the given values and solve for \(T_f\): \(T_f = 22.0 + \frac{(1.0010 / 1.0000) - 1}{19 \times 10^{-6}}\) \(T_f \approx 174.17^{\circ}\,\mathrm{C}\)
05

Conclusion

The brass ring must be heated to approximately 174.17°C so that the steel sphere at 22.0°C can just slip through.

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