Question

Ice starts forming in a lake with water at ${{0}^{0}}C$ when the atmospheric temperature is $-{{10}^{0}}C$ . If the time taken for the first 1cm of ice to be formed is 7 hours, then the time taken for the thickness of ice to change from 1cm to 2cm is:
(A) 7 hours
(B) 14 hours
(C) 21 hours
(D) 3.5 hours

Answer 1

Hint: We will first establish a relation between the time it takes to melt a certain layer of ice with the length of the ice layer. This can be done by calculating the amount of heat required to melt a very small layer of ice in a very small period of time, and then integrate the equation over certain limits.

Complete step-by-step answer:
Let us first assign some terms that we are going to use in our problem.
Let us say ‘x’ length of ice has already formed and it takes another ‘dt’ time to form ‘dx’ layer of ice over a surface area of ‘A’ with thermal conductivity of ice being ‘K’.
Also, let the mass of ice be ‘m’, its density ‘$\rho $’ and thus the volume of ice formed will be ‘Ax’.
Now, we can write the heat transferred as:
$\Rightarrow \dfrac{dQ}{dt}=\dfrac{KA(\vartriangle T)}{dx}$
Which can also be written as:
$\begin{align}
  & \Rightarrow \dfrac{d(mL)}{dt}=\dfrac{KA(\vartriangle T)}{dx} \\
 & \Rightarrow \dfrac{d(\rho AxL)}{dt}=\dfrac{KA(\vartriangle T)}{dx} \\
 & \Rightarrow dt=\dfrac{\rho L}{K\vartriangle T}xdx \\
\end{align}$
Where, L is the Latent heat of ice fusion of ice.
In the above equation, let all the constant terms collectively be denoted by ‘C’, then our equation becomes:
$\Rightarrow dt=Cxdx$
Integrating the above equation in limits ($0\to t$) for time and ($0\to l$) for length of ice formed, we get:
$\begin{align}
  & \Rightarrow \int\limits_{0}^{t}{dt}=\int\limits_{0}^{l}{Cxdx} \\
 & \Rightarrow t=C\dfrac{{{l}^{2}}}{2} \\
 & \therefore \dfrac{t}{{{l}^{2}}}=\text{constant} \\
\end{align}$
Therefore, we can write for two different layers of ice formed in two different amount of time as:
$\Rightarrow \dfrac{{{t}_{1}}}{l_{1}^{2}}=\dfrac{{{t}_{2}}}{l_{2}^{2}}$
Using the data given in our question, we get:
 $\begin{align}
  & \Rightarrow \dfrac{7}{{{1}^{2}}}=\dfrac{{{t}_{2}}}{{{2}^{2}}} \\
 & \therefore {{t}_{2}}=28hours \\
\end{align}$
Here, 28 hours is the time taken to form 2cm of ice from the start and 7 hours is the time taken to form 1cm of ice from the start. Thus to form the second 1cm of ice, the time taken is:
$\begin{align}
  & ={{t}_{2}}-{{t}_{1}} \\
 & =28-7 \\
 & =21hours \\
\end{align}$
Hence, then the time taken for the thickness of ice to change from 1cm to 2cm is 21 hours.

So, the correct answer is “Option C”.

Note: In this type of problem we should always go for the integration approach. If we had solved this problem by directly applying the formula, then the final answer to form the ice of 2cm thickness would be 14 hours which is incorrect, as at every instant the length of ice formed is not constant, that is, water is not freezing instantaneously but over a certain time.