Question

Ice starts forming in lake with water at $0^{\circ }$. When the atmospheric temperature is $-{10}^{\circ }C,$ if the time taken to form $1\text{cm}$ of ice be $7$ hours, then the time taken for the thickness of ice to change from $1\text{cm}$ to $2\text{cm}$ is

• A. $7$ hours
• B. $14$ hours
• C. $3.5$ hours
• D. $21$ hours

Answer 1

The correct option is D $21$ hours

Let us assume that $x$ length of ice has formed and to form a small strip of ice of length $dx$, $dt$ time is required, then $\frac{dQ}{dt}=\frac{KA\Delta T}{x}$
$\frac{d\left(mL\right)}{dt}=\frac{KA\Delta T}{x}$
$\frac{A{\rho }_{w}Ldx}{dt}=\frac{KA\Delta T}{x}$
$dt=\frac{{\rho }_{x}Lxdx}{K\Delta T}$
where,
$K$ is thermal conductivity of ice,
$A$ is area of the lake,
$L$ is talent heat of ice,
${\rho }_w$ are mass density of ice and
$\Delta T$ is the temperature difference.
By integrating the above equation on both, sides
${\int }_0^{t}dt=\frac{{\rho }_{w}L}{K\Delta T}{\int }_0^{l}xdx$
$t=\frac{{\rho }_{w}L}{K\Delta T}\frac{l^2}{2}$
$t=C\frac{l^2}{2}$
where $C=\frac{{\rho }_{w}L}{K\Delta T}$
Given that to form ice of 1 cm thickness, time of 7 hours has been taken then, time required to form 2 cm thickness of ice will be given by
$\frac{t_1}{l_1^2}=\frac{t^2}{l_2^2}$
$t_2=\frac{t_1}{l_1^2}{l}_2^2=\frac{7}{1}\times 4=28$ hours
Time taken to increase thickness of ice from $1\text{cm}$ to $2\text{cm}$ is $t_2-t_1=28-7=21$ hours