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Suppose you stand with one foot on ceramic flooring and one on a wool carpet, making contact over a 77.2 cm2 area with each foot. Both the ceramic and the carpet are 2.60 cm thick and are 10.0°C on their bottoms. At what rate must each foot supply heat to keep the top of the ceramic and carpet at 33.0°C? The thermal conductivity of ceramic is 0.84 J/(s · m · °C) and that of wool is

Respuesta :

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

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