Q. 24.0( 2 Votes )

# <span lang="EN-US

Answer :

**Given:**Thickness of the container : x = 1 cm = 0.01 m

Thermal conductivity of the sheet : K= 0.025 J s

^{–1}m

^{–1}°C

^{–1}

Temperature of the liquid nitrogen: T

_{1}= 80 K

Area of the container : A = 0.8 m

^{2}

Temperature of the atmosphere : T

_{2}= 300 K

**Formula used:**

Rate of amount of heat flowing is given as:

Here, Δθ is the amount of heat transferred, ΔT is the temperature difference, K is the thermal conductivity of the material, A is the area of cross section of the material and x is the thickness of the material.

Here, ΔT = T

_{2}- T

_{1}= 220 K

Substituting the values we get,

Hence, the rate of heat flow from the atmosphere to the liquid nitrogen is 440 J/s.

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