/*! 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 49 \(M\) The active element of a ce... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 49

\(M\) The active element of a certain laser is made of a glass rod \(30.0 \mathrm{~cm}\) long and \(1.50 \mathrm{~cm}\) in diameter. Assume the average coefficient of linear expansion of the glass is \(9.00 \times 10^{-6}\left({ }^{\circ} \mathrm{C}\right)^{-1}\). If the temperature of the rod increases by \(65.0^{\circ} \mathrm{C}\), what is the increase in (a) its length, (b) its diameter, and (c) its volume?

Short Answer

Expert verified
The increase in length is \( \approx 1.76 \times 10^{-2} \) m, the increase in diameter is \( \approx 0.0029 \times 10^{-2} \) m, and the increase in volume is \( \approx 2.22 \times 10^{-7} \) m鲁.

Step by step solution

01

Calculate the Increase in Length

The increase in length can be calculated using the formula for linear expansion: \( \Delta L = L_0 * \alpha * \Delta T \). Here, \( L_0 = 30.0 \) cm, \( \alpha = 9.00 \times 10^{-6} \degreeC^{-1} \), and \( \Delta T = 65.0 \degreeC \) . Substituting these values in, we get: \( \Delta L = 30.0 \times 10^{-2} m * 9.00 \times 10^{-6} \degreeC^{-1} * 65.0 \degreeC \).
02

Calculate the Increase in Diameter

Similar to Step 1, the increase in diameter can also be calculated using the same formula for linear expansion: \( \Delta d = d_0 * \alpha * \Delta T \). Here, \( d_0 = 1.50 \) cm, \( \alpha = 9.00 \times 10^{-6} \degreeC^{-1} \), and \( \Delta T = 65.0 \degreeC \) . Substituting these values in, we get: \( \Delta d = 1.50 \times 10^{-2} m * 9.00 \times 10^{-6} \degreeC^{-1} * 65.0 \degreeC \).
03

Calculate the Increase in Volume

The increase in volume can be calculated using the formula for volumetric expansion: \( \Delta V = V_0 * \beta * \Delta T \). However, we don't have the coefficient of volumetric expansion, but we know that \( \beta = 3 \alpha \) for most substances. So, we substitute this and rearrange the formula to \( \Delta V = V_0 * 3 * \alpha * \Delta T \). The original volume can be calculated using \( V_0 = \pi \times (d_0/2)^2 \times L_0 \). After substituting the values and calculating, we have \( \Delta V = \pi \times (1.50 \times 10^{-2} m/2)^2 \times 30.0 \times 10^{-2} m * 3 * 9.00 \times 10^{-6} \degreeC^{-1} * 65.0 \degreeC \).

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Most popular questions from this chapter

An expandable cylinder has its top connected to a spring with force constant \(2.00 \times 10^{3} \mathrm{~N} / \mathrm{m}\) (Fig. P10.60). The cylinder is filled with \(5.00 \mathrm{~L}\) of gas with the spring relaxed at a pressure of \(1.00 \mathrm{~atm}\) and a temperature of \(20.0^{\circ} \mathrm{C}\). (a) If the lid has a crosssectional area of \(0.0100 \mathrm{~m}^{2}\) and negligible mass, how high will the lid rise when the temperature is raised to \(250^{\circ} \mathrm{C}\) ? (b) What is the pressure of the gas at \(250^{\circ} \mathrm{C}\) ?

A brass ring of diameter \(10.00 \mathrm{~cm}\) at \(20.0^{\circ} \mathrm{C}\) is heated and slipped over an aluminum rod of diameter \(10.01 \mathrm{~cm}\) at \(20.0^{\circ} \mathrm{C}\). Assuming the average coefficients of linear expansion are constant, (a) to what temperature must the combination be cooled to separate the two metals? Is that temperature attainable? (b) What if the aluminum rod were \(10.02 \mathrm{~cm}\) in diameter?

At what temperature would the rms speed of helium atoms equal (a) the escape speed from Earth, \(1.12 \times 10^{4} \mathrm{~m} / \mathrm{s}\) and (b) the escape speed from the Moon, \(2.37 \times 10^{3} \mathrm{~m} / \mathrm{s}\) ? (See Chapter 7 for a discussion of escape speed.) Note: The mass of a helium atom is \(6.64 \times 10^{-27} \mathrm{~kg}\).

Gas is contained in an \(8.00\)-L vessel at a temperature of \(20.0^{\circ} \mathrm{C}\) and a pressure of \(9.00 \mathrm{~atm}\). (a) Determine the number of moles of gas in the vessel. (b) How many molecules are in the vessel?

The density of gasoline is \(7.30 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3}\) at \(0^{\circ} \mathrm{C}\). Its average coefficient of volume expansion is \(9.60 \times 10^{-4}\left({ }^{\circ} \mathrm{C}\right)^{-1}\), and note that \(1.00 \mathrm{gal}=0.00380 \mathrm{~m}^{3}\). (a) Calculate the mass of \(10.0 \mathrm{gal}\) of gas at \(0^{\circ} \mathrm{C}\). (b) If \(1.000 \mathrm{~m}^{3}\) of gasoline at \(0^{\circ} \mathrm{C}\) is warmed by \(20.0^{\circ} \mathrm{C}\), calculate its new volume. (c) Using the answer to part (b), calculate the density of gasoline at \(20.0^{\circ} \mathrm{C}\). (d) Calculate the mass of \(10.0\) gal of gas at \(20.0^{\circ} \mathrm{C}\). (e) How many extra kilograms of gasoline would you get if you bought \(10.0\) gal of gasoline at \(0^{\circ} \mathrm{C}\) rather than at \(20.0^{\circ} \mathrm{C}\) from a pump that is not temperature compensated?
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