#### Question

The normal body-temperature of a person is 97°F. Calculate the rate at which heat is flowing out of his body through the clothes assuming the following values. Room temperature = 47°F, surface of the body under clothes = 1.6 m^{2}, conductivity of the cloth = 0.04 J s^{−1} m^{−1}°C^{−1}, thickness of the cloth = 0.5 cm.

#### Solution

`"Rate of how of heat" = " tempreature diffrences" / "Thermal resistance"`

Temperature of the body, T_{1} = 97°F= 36. 11°C

Temperature of the surroundings, T_{2} = 47°F= 8.33°C`

Conductivity of the cloth, K = 0.04 J s^{-1 }m^{-1 °}c

Thickness of the cloth l = 0.5 cm = 0.005 m

Area of the cloth, A=1.6 m^{2}

Difference in the temperature ΔT = T_{1} - T_{2} = 36.11 - 8.33 = 27.78°C

Thermal resistance =` l / (KA) = 0.005/ ((0.04) × 1.6) =0.078125`

Rate at which heat is flowing out is given by

`(ΔQ)/(Δt) = (T_1 - T_2)/( l/(KA)`

`(ΔQ)/(Δt) = 27.78/ 0.078125`

`(ΔQ)/(Δt)` = 356 J / s