Question

A 9-ft-high room with a base area of \(12 \mathrm{ft} \times 12 \mathrm{ft}\) is to be heated by electric resistance heaters placed on the ceiling, which is maintained at a uniform temperature of \(90^{\circ} \mathrm{F}\) at all times. The floor of the room is at \(65^{\circ} \mathrm{F}\) and has an emissivity of \(0.8\). The side surfaces are well insulated. Treating the ceiling as a blackbody, determine the rate of heat loss from the room through the floor.

Short answer

Answer: The rate of heat loss from the room through the floor is approximately 1032.54 Watts.

Step-by-step solution

Identify constants and variables

First, let's list the given information: - Room dimensions: height of 9ft, base area of 12ft x 12ft - Ceiling temperature: \(T_c = 90^{\circ} F = (90 + 459.67) \,\mathrm {Rankine} = 549.67 \, \mathrm{R}\) (converted to Rankine for calculations) - Floor temperature: \(T_f = 65^{\circ} F = (65 + 459.67) \,\mathrm {Rankine} = 524.67 \, \mathrm{R}\) (converted to Rankine for calculations) - Emissivity of the floor: \(ε = 0.8\) - Ceiling treated as a blackbody (emissivity = 1)

Calculate the heat loss using the Stefan-Boltzmann Law

We need to find the heat loss rate through the floor using the Stefan-Boltzmann Law: $$q = εσA(T_c^4 - T_f^4)$$ where: - \(q\) is the heat loss rate - \(ε\) is the emissivity - \(σ\) is the Stefan-Boltzmann constant, \(5.67 \times 10^{-8} \, \frac {\mathrm{W}} {\mathrm{m^2} \cdot \mathrm{K^4}}\) - \(A\) is the area of the floor - \(T_c\) is the temperature of the ceiling in Rankine - \(T_f\) is the temperature of the floor in Rankine

Convert the area to metric units

Since we are using metric units for our Stefan-Boltzmann constant, we need to convert the base area of the room to metric units: $$A = 12 \,\mathrm{ft} × 12 \,\mathrm{ft} = 144\, \mathrm{ft^2} = 13.38 \,\mathrm{m^2}$$

Calculate the heat loss rate

Now we can plug in the values we found in Steps 1-3 into the Stefan-Boltzmann Law and solve for \(q\): $$q = 0.8 \times 5.67 \times 10^{-8}\, \frac {\mathrm{W}} {\mathrm{m^2} \cdot \mathrm{K^4}} \times 13.38\, \mathrm{m^2}((549.67\, \mathrm{R})^4 - (524.67\, \mathrm{R})^4)$$ After calculating, we find: $$q \approx 1032.54 \, \mathrm{W}$$ So, the rate of heat loss from the room through the floor is approximately 1032.54 Watts.