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

= 5.74W

=0.27 W

Explanation:

d= 2.6cm =>0.026m (for both the ceramic and the carpet )

Thermal conductivity of wool ''= = =  0.04J/(sm °C)

Thermal conducticity of carpet '' = 0.84 J/(sm °C)

Area 'A'= = = 77.2cm²=> 77.2 x m²

=33.0°C

=10.0°C

Average Power is determined by dividing amount of energy'Q' by time taken for the transfer't':

= Q/Δt

Due to conductivity, heat of flow rate will be P= dQ/dt

P= =

For CERAMIC:

= => [0.84 x 77.2 x (33-10) ]/0.026

= 5.74W

For WOOL CARPET:

= => [0.04 x 77.2 x (33-10) ]/0.026

=0.27 W

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