Question

A 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.

Answer 1

$\mathrm{Rate}\mathrm{of}\mathrm{flow}\mathrm{of}\mathrm{heat}=\frac{\mathrm{Temperature}\mathrm{difference}}{\mathrm{Thermal}\mathrm{resistance}}$
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, $\frac{l}{k\mathit{}A}=\frac{{10}^{-2}}{{\displaystyle 0.025\times 0.80}}$
Temperature difference, $∆T=T_1-T_2=300-80=220\mathrm{K}$

Rate of flow of heat, $\left(\frac{\Delta Q}{\Delta t}\right)=\frac{T_{\mathit{1}}\mathit{-}{T}_{\mathit{2}}}{{\displaystyle \frac{l}{kA}}}$
$$\begin{gathered} \Rightarrow \left(\frac{\Delta Q}{\Delta t}\right)=\frac{220}{{10}^{-2}}\times 0.025\times 0.80 \\ \Rightarrow \left(\frac{\Delta Q}{\Delta t}\right)=440\mathrm{J}/s \end{gathered}$$