Chapter 14, Problem 113P

(a)

To determine

The temperature at the copper –steel interface.

Answer to Problem 113P

Temperature is $103.81°\text{C}$.

Explanation of Solution

Write the equation for thermal conductivity from Fourier’s law of heat conduction.

Write the relation between thermal conductivities of copper and steel.

${\wp }_{st}={\wp }_{Cu}$ (I)

Here, ${\wp }_{st}$ is the rate of heat flow of steel and ${\wp }_{Cu}$ is the rate of heat flow of copper.

Write the equation for ${\wp }_{st}$.

${\wp }_{st}={\kappa }_{st}A\frac{\Delta T_{st}}{d_{st}}$

Here, ${\kappa }_{st}$ is the thermal conductivity of steel, $A$ is the cross sectional area, $\Delta T_{st}$ is the change in temperature of steel, and $d_{st}$ is the diameter of steel.

Write the equation for ${\wp }_{Cu}$.

${\wp }_{Cu}={\kappa }_{Cu}A\frac{\Delta T_{Cu}}{d_{Cu}}$

Here, ${\kappa }_{Cu}$ is the thermal conductivity of copper, $\Delta T_{Cu}$ is the change in temperature of copper, and $d_{Cu}$ is the diameter of copper.

Rewrite equation (I) by substituting the above relations for ${\wp }_{st}$ and ${\wp }_{Cu}$.

$\begin{array}{c}{\kappa }_{st}A\frac{\Delta T_{st}}{d_{st}}={\kappa }_{Cu}A\frac{\Delta T_{Cu}}{d_{Cu}}\\ \Delta T_{st}=\frac{{\kappa }_{Cu}{d}_{st}}{{\kappa }_{st}{d}_{Cu}}\Delta T_{Cu}\end{array}$ (I)

Write the equation to find the temperature at the copper –steel interface.

$\Delta T_{st}+\Delta T_{Cu}=T_c-T_b$ (II)

Here, $T_c$ is the temperature of ceramic heating element and $T_b$ is temperature of boiling water.

Write the equation for temperature at the copper –steel interface.

$T=T_c-\Delta T_{Cu}$ (III)

Here, $T$ is the temperature at the copper –steel interface.

Conclusion:

Substitute $401\text{W}/\text{m}\cdot \text{K}$ for ${\kappa }_{Cu}$, $46.0\text{W}/\text{m}\cdot \text{K}$ for ${\kappa }_{st}$, $0.350\text{cm}$ for $d_{st}$, and $0.150\text{cm}$ for $d_{Cu}$ in equation (I).

$\begin{array}{c}\Delta T_{st}=\frac{\left(401\text{W}/\text{m}\cdot \text{K}\right)\left(0.350\text{cm}\right)}{\left(46.0\text{W}/\text{m}\cdot \text{K}\right)\left(0.150\text{cm}\right)}\Delta T_{Cu}\\ =\left(20.3\right)\Delta T_{Cu}\end{array}$

Substitute $104.00°\text{C}$ for $T_c$, $100.00°\text{C}$ for $T_b$, and $\left(20.3\right)\Delta T_{Cu}$ for $\Delta T_{st}$ in equation (II) to find $\Delta T_{Cu}$.

$\begin{array}{c}\Delta T_{st}+\left(20.3\right)\Delta T_{Cu}=\left(104.00°\text{C}-100.00°\text{C}\right)\\ \Delta T_{st}=0.187°\text{C}\end{array}$

Substitute $104.00°\text{C}$ for $T_c$ and $0.187°\text{C}$ for $\Delta T_{st}$ in equation (III)

$\begin{array}{c}T=104.00°\text{C}-0.187°\text{C}\\ \text{=}103.81°\text{C}\end{array}$

Therefore, the temperature is $103.81°\text{C}$.

(b)

To determine

The rate of evaporation of water from the pan.

Answer to Problem 113P

Water evaporates at the rate of $0.565\text{g}/\text{s}$.

Explanation of Solution

Write the equation for heat flow.

$\frac{\Delta Q}{\Delta t}={\kappa }_{st}A\frac{\Delta T_{st}}{d_{st}}$

Here, $\Delta Q$ is the heat energy enters into the water and $\Delta t$ is the time interval.

Write the equation for $\Delta Q$.

$\Delta Q=\Delta m{L}_v$

Here,$\Delta m$ is the mass of water vapour and $L_v$ is the latent heat of vaporisation of water.

Rewrite the heat flow equation by substituting the above relation for $\Delta Q$.

$\begin{array}{c}\frac{\Delta m{L}_v}{\Delta t}={\kappa }_{st}A\frac{\Delta T_{st}}{d_{st}}\\ \frac{\Delta m}{\Delta t}=\frac{{\kappa }_{st}A\Delta T_{st}}{L_v{d}_{st}}\end{array}$

Conclusion:

Substitute $46.0\text{W}/\text{m}\cdot \text{K}$ for ${\kappa }_{st}$, $\pi {\left(\frac{0.180\text{m}}{2}\right)}^2$ for $A$, $103.81°\text{C}-100.00°\text{C}$ for $\Delta T_{st}$, $2256\text{J}/\text{g}$ for $L_v$, and $0.350\text{cm}$ for $d_{cm}$ in the above equation to find $\frac{\Delta m}{\Delta t}$.

$\begin{array}{c}\frac{\Delta m}{\Delta t}=\frac{\left(46.0\text{W}/\text{m}\cdot \text{K}\right)\left(\pi {\left(\frac{0.180\text{m}}{2}\right)}^2\right)\left(103.81°\text{C}-100.00°\text{C}\right)}{\left(2256\text{J}/\text{g}\right)\left(0.350\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}\\ =\left(5.65\times {10}^{-4}\text{kg}/\text{s}\right)\left(\frac{1\text{g}}{{10}^{-3}\text{kg}}\right)\\ =0.565\text{g}/\text{s}\end{array}$

Therefore, the water evaporates at the rate of $0.565\text{g}/\text{s}$.