NEET-XII-Physics
28: Heat Transfer
- #2A liquid-nitrogen container is made of a 1 cm thick styrofoam sheet having thermal conductivity 0.025 J s-1 m-1 °C-1. Liquid nitrogen at 80 K is kept in it. A total area of 0.80 m2 is in contact with the liquid nitrogen. The atmospheric temperature us 300 K. Calculate the rate of heat flow from the atmosphere to the liquid nitrogen.Ans : `` \,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,flow\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}=\frac{\,\mathrm{\,Temperature\,}\,\mathrm{\,difference\,}}{\,\mathrm{\,Thermal\,}\,\mathrm{\,resistance\,}}``
Thickness of the container, l = 1 cm = 10`` -``2 m
Thermal conductivity of the styrofoam sheet, `` k=0.025\,\mathrm{\,J\,}{\,\mathrm{\,s\,}}^{-1}{\,\mathrm{\,m\,}}^{-1}°{\,\mathrm{\,C\,}}^{-1}``
Area, A = 0.80 m2
Thermal resistance, `` \frac{l}{kA}=\frac{{10}^{-2}}{{\displaystyle 0.025\times 0.80}}``
Temperature difference, `` ∆T={T}_{1}-{T}_{2}=300-80=220\,\mathrm{\,K\,}``
Rate of flow of heat, `` \left(\frac{\Delta Q}{\Delta t}\right)=\frac{{T}_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}}{{\displaystyle \frac{l}{kA}}}``
`` \Rightarrow \left(\frac{\Delta Q}{\Delta t}\right)=\frac{220}{{10}^{-2}}\times 0.025\times 0.80``
`` \Rightarrow \left(\frac{\Delta Q}{\Delta t}\right)=440\,\mathrm{\,J\,}/s``
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