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5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 anddensity ρ = 0.9 × 103 kgm-3. Take thestep by step please
There is a formula for freezing of a lake.I am writing it down here.If you need to know how that formula came ask me here.Time taken=(dL/2KT)×(x22 - x12 )where d=density L=latent heatK=thermal conductivityT= temperature in degree centigradex1 =initial thicknes of ice sheet formedx2 = final thickness of ice sheet formed. In the given problem,T=|263-273|=10x1 =0 and x2 =0.5 meter and you know d,L,K.
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