NEET-XII-Physics
28: Heat Transfer
- #3The 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.Ans : `` \,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,flow\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}=\frac{\,\mathrm{\,Temperature\,}\,\mathrm{\,difference\,}}{\,\mathrm{\,Thermal\,}\,\mathrm{\,resistance\,}}``
Temperature of the body, `` {T}_{1}=97°\,\mathrm{\,F\,}=36.11°\,\mathrm{\,C\,}``
Temperature of the surroundings, `` {T}_{2}=47°\,\mathrm{\,F\,}=8.33°\,\mathrm{\,C\,}``
Conductivity of the cloth, `` K=0.04\,\mathrm{\,J\,}{\,\mathrm{\,s\,}}^{-1}{\,\mathrm{\,m\,}}^{-1}°{\,\mathrm{\,C\,}}^{-1}``
Thickness of the cloth, `` l=0.5\,\mathrm{\,cm\,}=0.005\,\mathrm{\,m\,}``
Area of the cloth, `` A=1.6{\,\mathrm{\,m\,}}^{2}``
Difference in the temperature, `` ∆T={T}_{1}-{T}_{2}=36.11-8.33=27.78°\,\mathrm{\,C\,}``
Thermal resistance = `` \frac{l}{KA}=\frac{0.005}{\left(0.04\right)\times 1.6}`` = 0.078125
Rate at which heat is flowing out is given by
`` \frac{\Delta Q}{\Delta t}=\frac{{T}_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}}{{\displaystyle \frac{l}{KA}}}``
`` \frac{\Delta Q}{\Delta t}=\frac{27.78}{0.078125}``
`` \frac{\Delta Q}{\Delta t}=356\,\mathrm{\,J\,}/s``
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