Question

The atmospheric temperature above a lake is below $0^o$C and constant. It is found that a $2$ cm layer of ice is formed in four days. In how many days will the thickness increase to $3$ cm?

• A. 9 days
• B. 5 days
• C. 7 days
• D. 6 days

Answer 1

Answer: (B)

5 days

Let $A$ be the area of lake.

$K$ = thermal conductivity of ice

$\rho$ = density of ice

$L$ = specific latent heat of ice

$-\theta$ = atmospheric temperature

If $x$ is the thickness of ice at any time $t$.

In additional time $dt$ let $dQ$ heat be conducted through ice upward from water at $0^{\circ }C$

$dQ=\frac{K\cdot A\cdot \cdot (0-(-\theta \left)\right)}{x}$ dt

If $dx$ is the increase in thickness of ice, the mass of ice formed$=Adx\rho$

$dQ=\left(A\rho dx\right)L$

$\frac{KA\theta }{x}dt=A\rho Ldx$

$\frac{K\theta }{\rho L}dt=xdx$

$\frac{K\theta }{\rho L}=constant\left(C\right)$

$Cdt=xdx$

$\int Cdt=\int xdx$

$Ct=\frac{x^2}{2}+constant$

At $t=0,x=0$

So that the constant = 0

$Ct=\frac{x^2}{2}$

$x^2\propto t$

$\frac{t}{4}=\frac{3^2}{2^2}$

$t=9days$

So the thickness increases from 2 cm to 3 cm is (9 - 4) = 5 days