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प्रश्न
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.
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उत्तर
`Rate of flow of heat = \text{Temperature diffrences}/ (Thermal \text{ resistance })`
Thickness of the container, l = 1 cm = 10 -2 m
Thermal conductivity of the styrofoam sheet, k = 0.025 J s-1 m-1 °C -1
Area, A= 0.80 m2
Thermal resistance , `l/ (KA) = 10^2/(0.025 xx 0.80)`
Temperature difference , ΔT = T1 - T2 = 300 - 80 = 220K
Rate of flow of heat, `( (ΔQ )/ (Δt) ) = (T1 - T2)/(l/KA)`
⇒ `( (ΔQ )/ (Δt) )` = 440 J / s
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