A cool box holds 2.0 kg of ice at 0 °C on a beach where the air is at 35 °C…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A cool box holds $2.0\ \text{kg}$ of ice at $0\ ^\circ\text{C}$ on a beach where the air is at $35\ ^\circ\text{C}$ . Its walls and lid are made of foam, $2.5\ \text{cm}$ thick, of thermal conductivity $0.035\ \text{W/(m K)}$ ; the foam itself (not counting the window) has a total area of $0.75\ \text{m}^2$ . The lid also has a small glass window, of area $0.050\ \text{m}^2$ , made of glass $4.0\ \text{mm}$ thick, of thermal conductivity $0.80\ \text{W/(m K)}$ , as shown in the figure. Take the inner surfaces to be at $0\ ^\circ\text{C}$ and the outer ones at $35\ ^\circ\text{C}$ , and ignore the corners. The latent heat of fusion of ice is $3.36 \times 10^5\ \text{J/kg}$ .

A cool box with foam walls 2.5 cm thick, holding ice. Its lid has a small glass window 4.0 mm thick. The inside is at 0 degrees Celsius and the air outside at 35 degrees Celsius.

At what rate does heat flow in through the foam, and through the window?
How long does it take for all the ice to melt?
A foam pad, $2.5\ \text{cm}$ thick and of the same foam, is now laid on top of the window, covering it exactly. How long does the ice last now?
With the pad in place, what is the temperature of the glass where it meets the pad?

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Treat each slab as a thermal resistance $R = d/(kA)$ , with heat current $\Delta T / R$ . Side by side the currents add; one after another the resistances add.

Step 1. Foam and window (a). Foam: $R = \dfrac{0.025}{0.035 \times 0.75}\ \text{K/W} = 0.952\ \text{K/W}$ , so the current is $\dfrac{35}{0.952}\ \text{W} = 36.75\ \text{W}$ . Window: $R = \dfrac{0.004}{0.80 \times 0.050}\ \text{K/W} = 0.10\ \text{K/W}$ , so the current is $350\ \text{W}$ . The small window lets in $90\%$ of the heat.

Step 2. Melting (b). Melting the ice needs $2.0 \times 3.36 \times 10^5\ \text{J} = 672000\ \text{J}$ . At $386.75\ \text{W}$ this takes $1738\ \text{s} = 29.0\ \text{min}$ .

Step 3. With the pad (c). The pad over the window has $R = \dfrac{0.025}{0.035 \times 0.050}\ \text{K/W} = 14.29\ \text{K/W}$ . In series with the glass, $R = 14.39\ \text{K/W}$ , so only $2.43\ \text{W}$ comes through the window. In all, $39.18\ \text{W}$ comes in, and the ice lasts $\dfrac{672000\ \text{J}}{39.18\ \text{W}} = 17150\ \text{s} = 4.76\ \text{h}$ .

Step 4. The glass (d). The same $2.43\ \text{W}$ crosses the glass, so the temperature across it rises by $2.43 \times 0.10 = 0.24\ \text{K}$ : the glass under the pad is at $0.24\ ^\circ\text{C}$ . Almost all the temperature drop is across the pad.

Answer: (a) $36.75\ \text{W}$ through the foam, $350\ \text{W}$ through the window. (b) $29.0\ \text{min}$ . (c) $4.76\ \text{h}$ . (d) $0.24\ ^\circ\text{C}$ .

Check. Glass conducts $23$ times better than foam, and the window is much thinner than the walls. So it is no surprise that a window with $\tfrac{1}{15}$ of the foam's area lets in most of the heat. Covering it makes the ice last $9.9$ times as long.

Watch out. Do not average the conductivities. Slabs side by side share the temperature difference and add their currents; slabs one after another share the current and add their resistances.

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