/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 86 An aluminium sphere of \(20 \mat... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 86

An aluminium sphere of \(20 \mathrm{~cm}\) diameter is heated from \(0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\). Its volume changes by (given that the coefficient of linear expansion for aluminium \(\left.\alpha_{A l}=23 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right)\) (A) \(28.9 \mathrm{cc}\) (B) \(2.89 \mathrm{cc}\) (C) \(9.28 \mathrm{cc}\) (D) \(49.8 \mathrm{cc}\)

Short Answer

Expert verified
The change in volume of the aluminium sphere when heated from \(0^{\circ}\) C to \(100^{\circ}\) C is approximately \(28.9 \mathrm{cm^3}\) (Option A).

Step by step solution

01

Find the initial volume of the sphere

To find the initial volume of the sphere, use the formula for the volume of a sphere, which is given by: \[V = \frac{4}{3}\pi r^3\] The diameter of the sphere is 20 cm, so the radius (r) is 10 cm. Plug the value of r into the formula and calculate the initial volume of the sphere: \[V = \frac{4}{3}\pi (10 \mathrm{cm})^3\] \[V = \frac{4000}{3}\pi \mathrm{cm^3}\]
02

Calculate the volume expansion factor

Use the formula for volume expansion, which is given by: \[V_{final} = V_{initial}(1+3\alpha \Delta T)\] Where \(\alpha\) is the coefficient of linear expansion, \(\Delta T\) is the change in temperature, and \(3\alpha \Delta T\) is the volume expansion factor. With the given values, find the volume expansion factor: \[3\alpha \Delta T = 3(23 \times 10^{-6}/{}^{\circ}C)(100^{\circ}C)\] \[3\alpha \Delta T = 6.9 \times 10^{-3}\]
03

Calculate the final volume of the sphere

Use the volume expansion factor to find the final volume of the sphere: \[V_{final} = V_{initial}(1+3\alpha \Delta T)\] \[V_{final} = \frac{4000}{3}\pi \mathrm{cm^3}(1+6.9\times10^{-3})\] \[V_{final} = \frac{4000}{3}\pi \mathrm{cm^3}(1.0069)\]
04

Determine the change in volume

Now, determine the change in volume by subtracting the initial volume from the final volume: \[\Delta V = V_{final} - V_{initial}\] \[\Delta V = (\frac{4000}{3}\pi \mathrm{cm^3})(1.0069) - \frac{4000}{3}\pi \mathrm{cm^3}\] \[\Delta V = \frac{4000}{3}\pi (0.0069)\mathrm{cm^3}\] Now, calculate the value of \(\Delta V\): \[\Delta V \approx 28.9 \mathrm{cm^3}\]
05

Choose the correct option

Compare the calculated value of \(\Delta V\) with the given options and choose the correct answer: (A) 28.9 cc is the correct answer

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Most popular questions from this chapter

A copper plate of length \(1 \mathrm{~m}\) is riveted to two steel plates of same length and same cross-section area at \(0^{\circ} \mathrm{C}\). Calculate tension (in kilo newton) generated in copper plate when heated to \(20^{\circ} \mathrm{C}\). \(Y_{\text {copper }}=\frac{1}{2} \times Y_{\text {steel }}=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2} Y=\) Young's modules \(\alpha_{\text {copper }}=18 \times 10^{-6} \mathrm{~K}^{-1}\) \(\alpha_{\text {steel }}=11 \times 10^{-6} \mathrm{~K}^{-1} \alpha=\) coefficient of linear expansion Area of each plate \(=50 \mathrm{~cm}^{2}\)

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A vessel of volume \(V\) contains a mixture of 1 mole of hydrogen and 1 mole oxygen (both considered as ideal). Let \(f_{1}(v) d v\) denote the fraction of molecules with speed between \(v\) and \((v+d v)\) with \(f_{2}(v) d v\), similarly for oxygen. Then, (A) \(f_{1}(v)+f_{2}(v)=f(v)\) obeys the Maxwell's distribution law. (B) \(f_{1}(v), f_{2}(v)\) will obey the Maxwell's distribution law separately. (C) neither \(f_{1}(v)\) nor \(f_{2}(v)\) will obey Maxwell's distribution law. (D) \(f_{2}(v)\) and \(f_{1}(v)\) will be the same.
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