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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 m2, 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, T1 = 97°F= 36. 11°C
Temperature of the surroundings, T2 = 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 m2
Difference in the temperature ΔT = T1 - T2 = 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
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