Calculating the Temperature Difference Maintained by a Heat Transfer: Conduction Through an Aluminum Pan

Conduction through the aluminum is the primary method of heat transfer here, and so we use the equation for the rate of heat transfer and solve for the temperature difference_.

1. Identify the knowns and convert them to the SI units.

   The thickness of the pan, $d=0\text{.800 cm}=8.0\times {\text{10}}^{-3}\text{ m,}$ the area of the pan, $A=\pi (0\text{.14}/2{)}^2{\text{ m}}^2=1\text{.}\text{54}\times {\text{10}}^{-2}{\text{ m}}^2$, and the thermal conductivity, $k=\text{220 J/s}\cdot \mathrm{m\cdot °}\mathrm{C.}$

2. Calculate the necessary heat of vaporization of 1 g of water:
   14.33
3. Calculate the rate of heat transfer given that 1 g of water melts in one second:
   $$Q/t=\text{2256}\mathrm{<mtext> J/s\; or\; 2.26\; kW.</mtext>}$$

   14.34
4. Insert the knowns into the equation and solve for the temperature difference:
   $$T_2-T_1=\frac{Q}{t}\left(\frac{d}{\text{kA}}\right)=\left(\text{2256}\mathrm{<mtext> J/s</mtext>}\right)\frac{8\text{.}\text{00} \times {\text{10}}^{-3}\mathrm{<mtext>m</mtext>}}{\left(\text{220}\mathrm{<mrow><mrow><mtext> J/s</mtext><mo\; stretchy="false">\cdot </mo><mtext>m</mtext></mrow><mo\; stretchy="false">\cdot º</mo></mrow><mtext>C</mtext>}\right)\left(1\text{.}\text{54}\times {\text{10}}^{-2}\mathrm{<msup><mtext> m</mtext><mrow><mn>2</mn></mrow></msup>}\right)}=5\text{.}\mathrm{33º}\mathrm{C.}$$

   14.35

The value for the heat transfer is typical for an electric stove. This value gives a remarkably small temperature difference between the stove and the pan. Consider that the stove burner is red hot while the inside of the pan is nearly $\text{100ºC}$ because of its contact with boiling water. This contact effectively cools the bottom of the pan in spite of its proximity to the very hot stove burner. Aluminum is such a good conductor that it only takes this small temperature difference to produce a heat transfer of 2.26 kW into the pan.

Conduction is caused by the random motion of atoms and molecules. As such, it is an ineffective mechanism for heat transport over macroscopic distances and short time distances. Take, for example, the temperature on the Earth, which would be unbearably cold during the night and extremely hot during the day if heat transport in the atmosphere was to be only through conduction. In another example, car engines would overheat unless there was a more efficient way to remove excess heat from the pistons.