5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of
263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 and
density ρ = 0.9 × 103 kgm-3. Take the
step by step please

Question

5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of
263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 and
density ρ = 0.9 × 103 kgm-3. Take the

step by step please

Susmita · Answer

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)×(x 2 2 - x 1 2 ) where d=density L=latent heat K=thermal conductivity T= temperature in degree centigrade x 1 =initial thicknes of ice sheet formed x 2 = final thickness of ice sheet formed. In the given problem, T=|263-273|=10 x 1 =0 and x 2 =0.5 meter and you know d,L,K.

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5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of 263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 and density ρ = 0.9 × 103 kgm-3. Take the step by step please

5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of
263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 and
density ρ = 0.9 × 103 kgm-3. Take the
step by step please

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5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of 263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 and density ρ = 0.9 × 103 kgm-3. Take the step by step please

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

1 Answers

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5) A small pond of depth 0.5 m deep is exposed to a cold winter with outside temperature of
263 K. Thermal conductivity of ice is K = 2.2 Wm-1K-1, latent heat L = 3.4 × 105 Jkg-1 and
density ρ = 0.9 × 103 kgm-3. Take the
step by step please