Question 11.11: Staying Warm in the Arctic Goal Combine two layers of insula...
Staying Warm in the Arctic
Goal Combine two layers of insulation.
Problem An arctic explorer builds a wooden shelter out of wooden planks that are 1.0 \mathrm{~cm} thick. To improve the insulation, he covers the shelter with a layer of ice 3.2 \mathrm{~cm} thick. (a) Compute the R factors for the wooden planks and the ice. (b) If the temperature outside the shelter is -20.0^{\circ} \mathrm{C} and the temperature inside is 5.00^{\circ} \mathrm{C}, find the rate of energy loss through one of the walls, if the wall has dimensions 2.00 \mathrm{~m} by 2.00 \mathrm{~m}. (c) Find the temperature at the interface between the wood and the ice.
Strategy After finding the R values, substitute into Equation 11.9
{\frac{Q}{\Delta t}}={\frac{A(T_{h}-T_{c})}{\sum\limits_{i}R_{i}}} (11.9)
to get the rate of energy transfer. To answer part (c), use Equation 11.7
\mathscr{P}=k A\,{\frac{(T_{h}-T_{c})}{L}} (11.7)
for one of the layers, setting it equal to the rate found in part (b), solving for the temperature.
(a) Compute the R values using the data in Table 11.3.
Find the R value for the wooden wall:
R_{\text {wood }}=\frac{L_{\text {wood }}}{k_{\text {wood }}}=\frac{0.01 \mathrm{~m}}{0.10 \mathrm{~J} / \mathrm{s} \cdot \mathrm{m} \cdot{ }^{\circ} \mathrm{C}}=0.10 \mathrm{~m}^{2} \cdot \mathrm{s} \cdot{ }^{\circ} \mathrm{C} / \mathrm{J}
Find the R-value for the ice layer:
R_{\text {ice }}=\frac{L_{\text {ice }}}{k_{\text {ice }}}=\frac{0.032 \mathrm{~m}}{1.6 \mathrm{~J} / \mathrm{s} \cdot \mathrm{m} \cdot{ }^{\circ} \mathrm{C}}=0.020 \mathrm{~m}^{2} \cdot \mathrm{s} \cdot{ }^{\circ} \mathrm{C} / \mathrm{J}
(b) Find the rate of heat loss.
Apply Equation 11.9:
\begin{aligned} \mathscr{P} & =\frac{Q}{\Delta t}=\frac{A\left(T_{h}-T_{c}\right)}{\sum_{i} R_{i}} \\ & =\frac{\left(4.00 \mathrm{~m}^{2}\right)\left(5.00^{\circ} \mathrm{C}-\left(-20.0^{\circ} \mathrm{C}\right)\right)}{0.12 \mathrm{~m}^{2} \cdot \mathrm{s} \cdot{ }^{\circ} \mathrm{C} / \mathrm{J}} \\ \mathscr{P} & =830 \mathrm{~W} \end{aligned}
(c) Find the temperature in between the ice and wood.
Apply the equation of heat conduction to the wood:
\begin{aligned} & \frac{k_{\text {wood }} A\left(T_{h}-T_{c}\right)}{L}=\mathscr{P} \\ & \frac{\left(0.10 \mathrm{~J} / \mathrm{s} \cdot \mathrm{m} \cdot{ }^{\circ} \mathrm{C}\right)\left(4.00 \mathrm{~m}^{2}\right)\left(5.00^{\circ} \mathrm{C}-T\right)}{0.010 \mathrm{~m}}=830 \mathrm{~W} \end{aligned}
Solve for the unknown temperature:
T=-16^{\circ} \mathrm{C}
Remarks The outer side of the wooden wall and the inner surface of the ice must have the same temperature, and the rate of energy transfer through the ice must be the same as through the wooden wall. Using Equation 11.7 for ice instead of wood gives the same answer. This rate of energy transfer is only a modest improvement over the thousandwatt rate in Exercise 11.9. The choice of insulating material is important!
TABLE 11.3
Thermal Conductivities
| Substance | Thermal Conductivity ( J/s · m · °C) Metals (at 25°C) |
| Metals (at 25°C) | |
| Aluminum | 238 |
| Copper | 397 |
| Gold | 314 |
| Iron | 79.5 |
| Lead | 34.7 |
| Silver | 427 |
| Gases (at 20°C) | |
| Air | 0.023 4 |
| Helium | 0.138 |
| Hydrogen | 0.172 |
| Nitrogen | 0.023 4 |
| Oxygen | 0.023 8 |
| Nonmetals | |
| Asbestos | 0.25 |
| Concrete | 1.3 |
| Glass | 0.84 |
| Ice | 1.6 |
| Rubber | 0.2 |
| Water | 0.60 |
| Wood | 0.10 |