Question

A lake is covered with ice 2 cm thick. The temperature of ambient air is -15⁰C. Find the rate of thickening of ice. For ice K = $4 \times {10^{ - 4}}$$kcal{m^{ - 1}}{s^{ - 1}}{\left( {^ \circ C} \right)^{ - 1}}$, density =$0.9 \times {10^3}kg{m^{ - 3}}$ and latent heat of ice(L) = $80kcalk{g^{ - 1}}$.

Answer 1

Hint: The heat energy flowing per second through the ice is equal to the heat energy per second due to the change in the state from water to ice i.e. latent heat.

Complete step by step answer:
The surface of a lake is covered with ice of thickness (y) = 2cm (0.02m) and the temperature (T) of surrounding air is -15⁰C. Let the heat energy flowing through them per second be H.
$H = \dfrac{{dQ}}{{dt}}$ where Q is the heat energy
$H = \dfrac{{\dfrac{{d\left\{ {KA\left( {{T_2} - {T_1}} \right)t} \right\}}}{y}}}{{dt}}$=$\dfrac{{\dfrac{{KAd\left( {{T_2} - {T_1}} \right)dt}}{y}}}{{dt}}$ [where K is a coefficient of thermal conductivity of ice, A is area, ${T_2}$ and ${T_1}$ are final and initial temperature and dt is change in time]

$\Rightarrow H = \dfrac{{KA\left( {dT} \right)}}{y}$ [Since K and A remain constant] [eqn1]
The latent heat of energy (L) is defined as the heat energy evolved in changing the state of a certain mass of a substance i.e., $Q = mL$and let $\dfrac{{dm}}{{dt}}$ represents a change of mass of ice in time t.
$\dfrac{{dm}}{{dt}} = \dfrac{{d\left( {Ay \times \rho } \right)}}{{dt}}$ [ $mass = volume \times density\left( \rho \right)$ and $volume = area\left( A \right) \times thickness$ ]
$\Rightarrow \dfrac{{dm}}{{dt}} = A\rho \dfrac{{dy}}{{dt}}$ [ Since area remain same and the height of ice only increases]
Thus, $H = \dfrac{{dQ}}{{dt}} \Rightarrow \dfrac{{d\left( {mL} \right)}}{{dt}}$
$H = L\dfrac{{dm}}{{dt}}$
Now, we substitute the value of $\dfrac{{dm}}{{dt}}$ in the above equation
$H = LA\rho \left( {\dfrac{{dy}}{{dt}}} \right)\left[ {eqn2} \right]$
We equate eqn1 and eqn2,
$\dfrac{{KAdT}}{y} = LA\rho \dfrac{{dy}}{{dt}}$
$\Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{KAdT}}{{yLA\rho }}$
$\Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{KdT}}{{yL\rho }}$
$\Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{4 \times {{10}^{ - 4}}k \times \left( {0 - \left( { - 15} \right)} \right)}}{{0.02 \times 80k \times 0.9 \times {{10}^3}}}$ [Since final temperature becomes 0⁰C due to formation of ice]
$\Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{4 \times {{10}^{ - 4}} \times 15}}{{1.44 \times {{10}^3}}}$
$\Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{60 \times {{10}^{ - 4}}}}{{1.44 \times {{10}^3}}}$
$\Rightarrow \dfrac{{dy}}{{dt}} = 41.67 \times {10^{ - 7}} = 4.167 \times {10^{ - 6}}m/s$
Therefore, the rate of thickening of ice is $4.167 \times {10^{ - 6}}m/s$.

Note:The heat energy is proportional to area, change in temperature and inversely proportional to change in thickness of the substance. The water under the surface of ice changes to ice due to the presence of ice above it and latent heat of energy is the energy which is required in changing the state from water to ice.