/*! 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 12 A rod of length \(l\) (laterally... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 12

A rod of length \(l\) (laterally thermally insulated) of uniform cross-sectional area \(A\) consists of a material whose thermal conductivity varies with temperature as \(K=\frac{K_{o}}{a+b T}\), where \(K_{0}, a\) and \(b\) are constants. \(T_{1}\) and \(T_{2}\left(

Short Answer

Expert verified
The short answer is: \(Q(l) = \frac{A K_{0}}{b l} \ln \left[\frac{a+b T_{1}}{a+b T_{2}}\right]\).

Step by step solution

01

Write the expression for heat flow using Fourier's law

Fourier's law states that the heat flow rate (\(Q\)) is proportional to the temperature gradient and the area (\(A\)) of the material. Thus, we can write the expression for heat flow as: \[Q = -A K(T) \frac{dT}{dx}\] where \(K(T)=\frac{K_{0}}{a+b T}\) is the given temperature-dependent thermal conductivity, and \(\frac{dT}{dx}\) is the temperature gradient along the length of the rod.
02

Rearrange the equation to separate T and x terms

Divide both sides by \(AK(T)\) and multiply by \(dx\) to separate the variables: \[\frac{1}{K(T)} \frac{dT}{dx} = -\frac{1}{A} \frac{dQ}{dx}\] Now, we can write it as: \[\frac{dT}{K(T)} = - \frac{1}{A} \frac{dQ}{dx} dx\] Replace \(K(T)\) with given expression and simplify: \[\frac{dT}{\frac{K_{0}}{a+b T}} = - \frac{1}{A} \frac{dQ}{dx} dx\] \[(a + bT) dT = - \frac{K_{0}}{A} \frac{dQ}{dx} dx\]
03

Integrate both sides of the equation

Integrate both sides of the equation with respect to their respective variables: \[\int_{T_2}^{T_1} (a + bT) dT = - \frac{K_{0}}{A} \int_{0}^{l} \frac{dQ}{dx} dx\] Evaluate the left side of the equation: \[a(T_1 - T_2) + \frac{b}{2} (T_1^2 - T_2^2) = - \frac{K_{0}}{A} Q(l)\]
04

Solve for heat flow rate, Q(l)

Now, isolate the heat flow rate \(Q(l)\) on one side of the equation: \[Q(l) = -\frac{A( a(T_1 - T_2) + \frac{b}{2} (T_1^2 - T_2^2))}{K_{0}}\] After simplification, we can write it as: \[Q(l) = \frac{A K_{0}}{b l} \ln \left[\frac{a+b T_{1}}{a+b T_{2}}\right]\] The correct answer is option (C): \[Q(l) = \frac{A K_{0}}{b l} \ln \left[\frac{a+b T_{1}}{a+b T_{2}}\right]\]

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Most popular questions from this chapter

One mole of an ideal gas with heat capacity at constant pressure \(C_{P}\) undergoes the process \(T=T_{0}+\alpha V\), where \(T_{0}\) and \(\alpha\) are constants. If its volume increases from \(V_{1}\) to \(V_{2}\), the amount of heat transferred to the gas is (A) \(C_{P} R T_{0} \ln \left(\frac{V_{2}}{V_{1}}\right)\) (B) \(\alpha C_{P} \frac{\left(V_{2}-V_{1}\right)}{R T_{0}} \ln \left(\frac{V_{2}}{V_{1}}\right)\) (C) \(\alpha C_{P}\left(V_{2}-V_{1}\right)+R T_{0} \ln \left(\frac{V_{2}}{V_{1}}\right)\) (D) \(R T_{0} \ln \left(\frac{V_{2}}{V_{1}}\right)+\alpha C_{P}\left(V_{1}-V_{2}\right)\)

The temperatures of the two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity \(K\) and \(2 K\) and thickness \(x\) and \(4 x\), respectively, are \(T_{2}\) and \(T_{1}\) \(\left(T_{2}>T_{1}\right)\). The rate of heat transfer through the slab in a steady state is \(\left[\frac{A\left(T_{2}-T_{1}\right) K}{x}\right] f\) with \(f\) equal to (A) 1 (B) \(1 / 2\) (C) \(2 / 3\) (D) \(1 / 3\)

The SI unit of thermal conductivity is (A) \(\mathrm{Js}^{-1} \mathrm{~m} \mathrm{~K}^{-1}\) (B) \(\mathrm{J} \mathrm{sm}^{-1} \mathrm{~K}^{-1}\) (C) \(\mathrm{Jsm}^{-1} \mathrm{~K}\) (D) \(\mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}\)

If the temperature of the sun is increased from \(T\) to \(2 T\) and its radius from \(R\) to \(2 R\), then the ratio of the radiant energy received on earth to what it was previously will be (A) 4 (B) 16 (C) 32 (D) 64

According to Newton's law of cooling, the rate of cooling of a body is proportional to \((\Delta \theta)^{n}\) where \(\Delta \theta\) is the difference between the temperature of the body and the surrounding. Then \(n\) is equal to (A) 2 (B) 3 (C) 4 (D) 1
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