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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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Most popular questions from this chapter

A smoke particle has a mass of \(1.38 \times 10^{-17} \mathrm{kg}\) and it is randomly moving about in thermal equilibrium with room temperature air at \(27^{\circ} \mathrm{C} .\) What is the rms speed of the particle?
The driver from Practice Problem 13.3 fills his \(18.9-\mathrm{L}\) steel gasoline can in the morning when the temperature of the can and the gasoline is \(15.0^{\circ} \mathrm{C}\) and the pressure is 1.0 atm, but this time he remembers to replace the tightly fitting cap after filling the can. Assume that the can is completely full of gasoline (no air space) and that the cap does not leak. The temperature climbs to \(30.0^{\circ} \mathrm{C}\) Ignoring the expansion of the steel can, what would be the pressure of the heated gasoline? The bulk modulus for gasoline is $1.00 \times 10^{9} \mathrm{N} / \mathrm{m}^{2}$
The coefficient of linear expansion of brass is \(1.9 \times 10^{-5}\) $^{\circ} \mathrm{C}^{-1} .\( At \)20.0^{\circ} \mathrm{C},$ a hole in a sheet of brass has an area of \(1.00 \mathrm{mm}^{2} .\) How much larger is the area of the hole at $30.0^{\circ} \mathrm{C} ?(\text { Wile tutorial: loop around the equator })$
A cylinder with an interior cross-sectional area of \(70.0 \mathrm{cm}^{2}\) has a moveable piston of mass \(5.40 \mathrm{kg}\) at the top that can move up and down without friction. The cylinder contains $2.25 \times 10^{-3} \mathrm{mol}\( of an ideal gas at \)23.0^{\circ} \mathrm{C} .$ (a) What is the volume of the gas when the piston is in equilibrium? Assume the air pressure outside the cylinder is 1.00 atm. (b) By what factor does the volume change if the gas temperature is raised to \(223.0^{\circ} \mathrm{C}\) and the piston moves until it is again in equilibrium?
Find the molar mass of ammonia (NH \(_{3}\) ).
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