A layer of ice of 20 cm thickness accumulates on the surface of t… - Vidyadip

Question

A layer of ice of 20 cm thickness accumulates on the surface of the water of a pond. The temperature of air is $-10^{\circ} C$. The thickness of the ice layer is 0.1 cm, how long will the increase take? The conductivity of ice is $2.1 \text {Joule}~ \text {sec} ^{-1} m^{-1} k ^{-1}$, latent heat $3.36 \times 10^5 \text {Joule}~\text {kg} ^{-1}$ and density $10^3$.

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Answer

Given :
Latent heat of ice $= L =3.36 \times 10^5$ Joule $kg ^{-1}$
Thickness of surface of ice $=20 cm=0.20 m$
Temperature under the ice layer $=-10^{\circ} C$
0.1 cm on 1 meter surface of ice ( 0.001 m )
Volume of thick ice layer $=1 \times 0.001$
$=.001 m^3$
Mass of ice $.001 m^3= m =0.001 \times 10^3=1 kg$
Energy lost to convert 1 kg of water into ice
$\begin{array}{l}=mL \\=1 \times 3.36 \times 10^5 \text { Joule }\end{array}$
Radiated energy from ice $1 m^3=\frac{ KA \left( T _2- T _1\right) \times t}{d}$
$\begin{aligned}3 .36 \times 10^5 & =\frac{2.1 \times 1(0-(-10)) \times t}{0.2} \\t & =\frac{3.36 \times 10^5 \times 0.2}{21} \\& =\frac{672}{21} \times 10^2=32 \times 10^2 \\& = 3200~ \text {sec} .\end{aligned}$

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