Answer:
[/tex] 408.8[/tex]
Explanation:
[tex]t[/tex] = thickness of the floor = 0.159 m
[tex]k[/tex] = thermal conductivity of wooden floor = 0.141 Wm⁻¹ ⁰C⁻¹
[tex]T_{i}[/tex] = Temperature of the interior of the cabin = 18.0 ⁰C
[tex]T_{o}[/tex] = Temperature of the air below the cabin = - 16.4 ⁰C
Difference in temperature is given as
[tex]\Delta T[/tex] = Difference in temperature = [tex]T_{i} - T_{o} = 18 - (- 16.4) = 34.4[/tex]⁰C
[tex]A[/tex] = Area of the floor = 13.4 m²
[tex]Q[/tex] = Rate of heat conduction
Rate of heat conduction is given as
[tex]Q = \frac{kA \Delta T}{t}[/tex]
[tex]Q = \frac{(0.141) (13.4) (34.4)}{0.159}\\Q = 408.8[/tex] W