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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

Question Image
Grade:12

1 Answers

Susmita
425 Points
3 years ago
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 heat
K=thermal  conductivity
T= temperature in degree centigrade
x1 =initial thicknes of ice sheet formed
x2 = final  thickness of ice sheet formed.
 
In the given problem,
T=|263-273|=10
x=0  and x2 =0.5 meter and you know d,L,K.

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