Calculating Heat Transfer Through Conduction: Conduction Rate Through an Ice Box

This question involves both heat for a phase change (melting of ice) and the transfer of heat by conduction. To find the amount of ice melted, we must find the net heat transferred. This value can be obtained by calculating the rate of heat transfer by conduction and multiplying by time.

1. Identify the knowns.
   $$\begin{array}{c}A=0.950{\text{m}}^2\text{;}\\ d=2.50\text{cm}=\mathrm{0.0250\; m;}\\ T_1=\mathrm{0º}\text{C;}\\ T_2=\mathrm{35.0º}\text{C,}t=\text{1 day}=\text{24 hours}=\text{86,400 s.}\end{array}$$

   14.27
2. Identify the unknowns. We need to solve for the mass of the ice, $m$. We will also need to solve for the net heat transferred to melt the ice, $Q$.
3. Determine which equations to use. The rate of heat transfer by conduction is given by
   $$\frac{Q}{t}=\frac{\text{kA}(T_2-T_1)}{d}\text{.}$$

   14.28
4. The heat is used to melt the ice: $Q={\text{mL}}_{\text{f}}.$
5. Insert the known values:
   $$\frac{Q}{t}=\frac{\left(\text{0.010 J/s}\cdot \text{m}\cdot º\text{C}\right)\left(\text{0.950}{\text{ m}}^2\right)\left(\text{35.}\mathrm{0º}\text{C}-\mathrm{0º}\text{C}\right)}{\text{0.0250 m}}=\text{13.3 J/s.}$$

   14.29
6. Multiply the rate of heat transfer by the time ():
   $$Q=\left(Q/t\right)t=\left(\text{13}\text{.}3\mathrm{<mtext> J/s</mtext>}\right)\left(\text{86},\text{400}\mathrm{<mtext> s</mtext>}\right)=1\text{.}\text{15}\times {\text{10}}^6\mathrm{<mtext> J.</mtext>}$$

   14.30
7. Set this equal to the heat transferred to melt the ice: $Q={\text{mL}}_{\text{f}}$. Solve for the mass $m$:
   $$m=\frac{Q}{L_{\text{f}}}=\frac{1\text{.}\text{15}\times {\text{10}}^6\mathrm{<mtext> J</mtext>}}{\text{334 }\times {\text{10}}^3\mathrm{<mtext> J/kg</mtext>}}=3\text{.}\text{44}\mathrm{<mtext>kg.</mtext>}$$

   14.31

The result of 3.44 kg, or about 7.6 lbs, seems about right, based on experience. You might expect to use about a 4 kg (7–10 lb) bag of ice per day. A little extra ice is required if you add any warm food or beverages.

Inspecting the conductivities in Table 14.3 shows that Styrofoam is a very poor conductor and thus a good insulator. Other good insulators include fiberglass, wool, and goose-down feathers. Like Styrofoam, these all incorporate many small pockets of air, taking advantage of air’s poor thermal conductivity.