Answer:
[tex]P_{c[/tex]= 5.74W
[tex]P_{w}[/tex]=0.27 W
Explanation:
d= 2.6cm =>0.026m (for both the ceramic and the carpet )
Thermal conductivity of wool '[tex]k_{w}[/tex]'= [tex]k_{carpet}[/tex] = [tex]k_{wool}[/tex]= 0.04J/(sm °C)
Thermal conducticity of carpet '[tex]k_c}[/tex]' = 0.84 J/(sm °C)
Area 'A'= [tex]A_{carpet}[/tex]= [tex]A_{ceramic}[/tex]= 77.2cm²=> 77.2 x [tex]10^{-4}[/tex]m²
[tex]T_h}[/tex]=33.0°C
[tex]T_c}[/tex]=10.0°C
Average Power [tex]P_{avg}[/tex] is determined by dividing amount of energy'Q' by time taken for the transfer't':
[tex]P_{avg}[/tex] = Q/Δt
Due to conductivity, heat of flow rate will be P= dQ/dt
P=[tex]\frac{dQ}{dt}[/tex] = [tex]\frac{kA[T_{h}-T_{c} ]}{d}[/tex]
For CERAMIC:
[tex]P_{c[/tex]=[tex]\frac{k_{c} A[T_{h}-T_{c} ]}{d}[/tex] => [0.84 x 77.2 x [tex]10^{-4}[/tex](33-10) ]/0.026
[tex]P_{c[/tex]= 5.74W
For WOOL CARPET:
[tex]P_{w}[/tex]= [tex]\frac{k_{w} A[T_{h}-T_{c} ]}{d}[/tex]=> [0.04 x 77.2 x [tex]10^{-4}[/tex](33-10) ]/0.026
[tex]P_{w}[/tex]=0.27 W