Question

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

Answer 1

`P_{c`= 5.74W

`P_(w)`=0.27 W

Step-by-step explanation:

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

Thermal conductivity of wool '
`k_(w)`'=
`k_(carpet)` =
`k_(wool)`= 0.04J/(sm °C)

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

Area 'A'=
`A_(carpet)`=
`A_(ceramic)`= 77.2cm²=> 77.2 x
`10^(-4)`m²

`T_h}`=33.0°C

`T_c}`=10.0°C

Average Power
`P_(avg)` is determined by dividing amount of energy'Q' by time taken for the transfer't':

`P_(avg)` = Q/Δt

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

P=
`(dQ)/(dt)` =
`(kA[T_(h)-T_(c) ])/(d)`

For CERAMIC:

`P_{c`=
`(k_(c) A[T_(h)-T_(c) ])/(d)` => [0.84 x 77.2 x
`10^(-4)`(33-10) ]/0.026

`P_{c`= 5.74W

For WOOL CARPET:

`P_(w)`=
`(k_(w) A[T_(h)-T_(c) ])/(d)`=> [0.04 x 77.2 x
`10^(-4)`(33-10) ]/0.026

`P_(w)`=0.27 W