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The temperature of the water of a pond is $0^{\circ} \mathrm{C}$ while that of the surrounding atmosphere is $-20^{\circ} \mathrm{C}$. If the density of ice is $\rho$ coefficient of thermal conductivity is $k$ and latent heat of melting is $L,$ then the thickness $Z$ of ice layer formed increases as a function of time $t$ is
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The correct answer is:
$Z^{2}=\frac{40 k}{\rho L} t$
Temperature of water $=0^{\circ} \mathrm{C}$
Temperature of surrounding atmosphere $=-20^{\circ} \mathrm{C}$
Density of ice $=\rho$
Coefficient of thermal conductivity $=k$
latent heat of melting $=L$
The rate of heat flow $H=\frac{d Q}{d t}=\frac{k A\left(T_{1}-T_{2}\right)}{x}$
$$
\begin{array}{l}
=\frac{k f(0-(-20)]}{x} \\
=\frac{k A \times 20}{x} \\
\frac{d Q}{d t}=\frac{d m}{d t} \cdot L \quad(\because m L=Q) \\
d m=\rho A \cdot d x \\
\frac{d Q}{d t}=\rho A \cdot \frac{d x}{d t} L=\frac{k A \cdot 20}{x} \\
\int_{0}^{Z} x \cdot d x=\frac{20 k}{\rho L} \int_{0}^{t} d t \Rightarrow Z^{2}=\frac{40 k}{\rho L} t
\end{array}
$$
So, thickness $Z$ increases as a function of time $t$ according to the above equation.
Temperature of surrounding atmosphere $=-20^{\circ} \mathrm{C}$
Density of ice $=\rho$
Coefficient of thermal conductivity $=k$
latent heat of melting $=L$
The rate of heat flow $H=\frac{d Q}{d t}=\frac{k A\left(T_{1}-T_{2}\right)}{x}$
$$
\begin{array}{l}
=\frac{k f(0-(-20)]}{x} \\
=\frac{k A \times 20}{x} \\
\frac{d Q}{d t}=\frac{d m}{d t} \cdot L \quad(\because m L=Q) \\
d m=\rho A \cdot d x \\
\frac{d Q}{d t}=\rho A \cdot \frac{d x}{d t} L=\frac{k A \cdot 20}{x} \\
\int_{0}^{Z} x \cdot d x=\frac{20 k}{\rho L} \int_{0}^{t} d t \Rightarrow Z^{2}=\frac{40 k}{\rho L} t
\end{array}
$$
So, thickness $Z$ increases as a function of time $t$ according to the above equation.
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