Chapter 14, Problem 90P

(a)

To determine

The rate of heat conduction along the bar.

Answer to Problem 90P

The rate of heat conduction along the bar is $0.32\text{ W}$ .

Explanation of Solution

Write the Fourier’s law of heat conduction.

$\wp =\kappa A\frac{\Delta T}{d}$ (I)

Here, $\wp$ is the rate of heat conduction, $\kappa$ is the thermal conductivity of material of the bar, $A$ is the cross sectional area, $\Delta T$ is the change in temperature and $d$ is the length of the bar

Write the equation for $\Delta T$ .

$\Delta T=T_1-T_2$ (II)

Here, $T_1$ is the temperature of the first end of the bar and $T_2$ is the temperature of  the second end of the bar

Put equation (II) in equation (I).

$\wp =\kappa A\frac{T_1-T_2}{d}$ (III)

Conclusion:

Given that the cross sectional area of the bar is $1.0\times {10}^{-6}{\text{ m}}^2$ and its length is $0.10\text{ m}$ , the thermal conductivity of copper is $401\text{ W/m}\cdot \text{K}$ , temperature of the first end of the bar is $104°\text{C}$ and the temperature of the other end is $24°\text{C}$ .

Substitute $401\text{ W/m}\cdot \text{K}$ for $\kappa$ , $1.0\times {10}^{-6}{\text{ m}}^2$ for $A$ , $0.10\text{ m}$ for $d$ , $104°\text{C}$ for $T_1$ and $24°\text{C}$ for $T_2$ in equation (III) to find $\wp$ .

$\begin{array}{c}\wp =\left(401\text{ W/m}\cdot \text{K}\right)\left(1.0\times {10}^{-6}{\text{ m}}^2\right)\frac{104°\text{C}-24°\text{C}}{0.10\text{ m}}\\ =0.32\text{ W}\end{array}$

Therefore, the rate of heat conduction along the bar is $0.32\text{ W}$ .

(b)

To determine

The temperature gradient in the bar.

Answer to Problem 90P

The temperature gradient in the bar is $800\text{ K/m}$ .

Explanation of Solution

Write the equation for the temperature gradient.

$\text{temperature gradient}=\frac{\Delta T}{d}$

Put equation (II) in the above equation.

$\text{temperature gradient}=\frac{T_1-T_2}{d}$ (IV)

Conclusion:

Substitute $104°\text{C}$ for $T_1$ , $24°\text{C}$ for $T_2$ and $0.10\text{ m}$ for $d$ in equation (IV) to find the temperature gradient .

$\begin{array}{c}\text{temperature gradient}=\frac{104°\text{C}-24°\text{C}}{0.10\text{ m}}\\ =\frac{80\text{ K}}{0.10\text{ m}}\\ =800\text{ K/m}\end{array}$

Therefore, the temperature gradient in the bar is $800\text{ K/m}$ .

(c)

To determine

The rate of heat conduction if two bars are placed in series between the same temperature baths.

Answer to Problem 90P

The rate of heat conduction if two bars are placed in series between the same temperature baths is $0.16\text{ W}$ .

Explanation of Solution

When two similar bars are placed in series, the effective length will be doubled. Equation (III) can be used to find the rate of heat conduction with the value of length doubled.

Conclusion:

The new length is $0.20\text{ m}$ .

Substitute $401\text{ W/m}\cdot \text{K}$ for $\kappa$ , $1.0\times {10}^{-6}{\text{ m}}^2$ for $A$ , $0.20\text{ m}$ for $d$ , $104°\text{C}$ for $T_1$ and $24°\text{C}$ for $T_2$ in equation (III) to find $\wp$ .

$\begin{array}{c}\wp =\left(401\text{ W/m}\cdot \text{K}\right)\left(1.0\times {10}^{-6}{\text{ m}}^2\right)\frac{104°\text{C}-24°\text{C}}{0.20\text{ m}}\\ =0.16\text{ W}\end{array}$

Therefore, the rate of heat conduction if two bars are placed in series between the same temperature baths is $0.16\text{ W}$ .

(d)

To determine

The rate of heat conduction if two bars are placed in parallel between the same temperature baths.

Answer to Problem 90P

The rate of heat conduction if two bars are placed in parallel between the same temperature baths is $0.64\text{ W}$ .

Explanation of Solution

When the two identical bars are placed parallel, the effective area gets doubled. Equation (III) can be used to find the rate of heat conduction with the value of area doubled.

Conclusion:

The new area is $2.0\times {10}^{-6}{\text{ m}}^2$ .

Substitute $401\text{ W/m}\cdot \text{K}$ for $\kappa$ , $2.0\times {10}^{-6}{\text{ m}}^2$ for $A$ , $0.10\text{ m}$ for $d$ , $104°\text{C}$ for $T_1$ and $24°\text{C}$ for $T_2$ in equation (III) to find $\wp$ .

$\begin{array}{c}\wp =\left(401\text{ W/m}\cdot \text{K}\right)\left(2.0\times {10}^{-6}{\text{ m}}^2\right)\frac{104°\text{C}-24°\text{C}}{0.10\text{ m}}\\ =0.64\text{ W}\end{array}$

Therefore, the rate of heat conduction if two bars are placed in parallel between the same temperature baths is $0.64\text{ W}$ .

(e)

To determine

The temperature at the junction, where the bars meet, in series case.

Answer to Problem 90P

The temperature at the junction, where the bars meet, in series case is $64°\text{C}$ .

Explanation of Solution

The bars are identical. This implies the temperature at the junction will be midway between the temperatures of the constant-temperature paths.

Write the equation for the temperature at the junction.

$T=\frac{T_1+T_2}{2}$ (V

Here, $T$ is the temperature of the junction

Conclusion:

Substitute $104°\text{C}$ for $T_1$ and $24°\text{C}$ for $T_2$ in equation (V) to find $T$ .

$\begin{array}{c}\text{temperature gradient}=\frac{104°\text{C+}24°\text{C}}{2}\\ =\frac{128°\text{C}}{2}\\ =64°\text{C}\end{array}$

Therefore, the temperature at the junction, where the bars meet, in series case is $64°\text{C}$ .