The sun shines on a 150 m² road surface so it is at 45°C. Below the 5cm thick asphalt (average conductivity of 0.06 W/m K), is a layer of rubbles at 15°C. Find the rate of heat transfer to the rubbles. Select one: a. 5400 W b. 5500 W c. 5300 W d. 5600 W
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The sun shines on a $150-\mathrm{m}^{2}$ road surface so that it is at $45^{\circ} \mathrm{C}$. Below the 5 -cm-thick asphalt, with average conductivity of $0.06 \mathrm{W} / \mathrm{m} \mathrm{K},$ is a layer of compacted nubble at a temperature of $15^{\circ} \mathrm{C}$. Find the rate of heat transfer to the rubble.
The sun shines on a $1500-\mathrm{ft}^{2}$ road surface so it is at 115 F. Below the 2 -in. -thick asphalt, average conductivity of 0.035 Btu/h $\mathrm{A} \mathrm{F}$, is a layer of compacted rubble at a temperature of 60 F. Find the rate of heat transfer to the rubble.
Consider sunlight shining on the asphalt surface of a paved road. When the morning sun first shines on the asphalt, the cold asphalt begins to warm up. By noon, the asphalt is quite hot, and has reached a constant temperature. Suppose at noon the Sun delivers 1000 W of power to each square meter of asphalt. Based on this information, calculate the temperature of the asphalt at noon. Assume the emissivity of asphalt is e = 0.93. Also assume that the only mechanism for heating the asphalt is thermal radiation. In other words, energy transfer through thermal conduction and convection is negligible.
Sri K.
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