Question

Ice starts forming in a lake with water 0oC,

when atmospheric temperature is -10oC. If the

time taken for 5 mm thickness of ice to be

formed is 2 hours, then the time taken for the

thickness of ice to change from 5 mm to

20 mm, is​

Answer 1

$\huge\sf{\gray{\underbrace{\red{QUESTION:-}}}}$

⚡ Ice starts forming in a lake with water 0°C, when atmospheric temperature is -10°C . If the time taken for 5mm thickness of ice to be formed is 2 hours . Then find the time taken for the thickness of ice changes from 5mm to 20mm .

$\huge\sf{\gray{\underbrace{\red{ANSWER:-}}}}$

$\mathcal{\gray{\underbrace{\blue{GIVEN:-}}}}$

• Ice starts forming in a lake with water 0°C, when atmospheric temperature is -10°C .

• The time taken for 5mm thickness of ice to be formed is 2 hours .

$\mathcal{\gray{\underbrace{\blue{TO\: FIND:-}}}}$

• The time taken for the thickness of ice changes from 5mm to 20mm .

$\mathcal{\gray{\underbrace{\blue{DIAGRAM:-}}}}$

✍️ See the above diagram .

$\mathcal{\gray{\underbrace{\blue{CALCULATION:-}}}}$

Let,

⚡ At “t = t”, x thickness of ice is formed as shown in figure .

⚡ And at “t = (t + dt)”, dx thickness of ice is formed as shown in figure .

⚡ Take, area of the lake is “A” .

So,

$\rm{\green{Mass\:of\:water\:converted\:into\:ice\:{\atop{[in\:dt]}}\:=\:\bold{\rho\:A\:dx}\:}}$

$\rm{\implies\:dQ\:=\:\rho\:A\:dx\:}$

$\rm{\implies\:\dfrac{dQ}{dt}\:=\:\dfrac{\rho\:A\:dx}{dt}\:}$

Where,

• dQ = change in Heat .

✍️ This heat passes through the upper layer of the lake from atmosphere .

Hence, we can also write

$\rm{\implies\:\dfrac{dQ}{dt}\:=\:\dfrac{\rho\:A\:dx}{dt}\:=\:\dfrac{k\:A}{x}\times{10}\:}$

$\rm{\implies\:\dfrac{\rho\:A\:dx}{dt}\:=\:\dfrac{k\:A}{x}\times{10}\:}$

$\rm{\implies\:\dfrac{\rho}{k}\times\:\int\limits_{x_1}^{x_2}\:x\:.\:dx\:=\:10\times\:\int\limits_{t_1}^{t_2}\:dt\:}$

$\rm{\implies\:\dfrac{\rho}{k}\:(x_2^2\:-\:x_1^2)\:=\:10\:(t_2\:-\:t_1)\:}$

$\rm\green{\implies\:(t_2\:-\:t_1)\:=\:\dfrac{\rho}{10\:k}\:(x_2^2\:-\:x_1^2)\:}$

CASE - 1 :-

• $\rm{t_2\:=\:2h}$

• $\rm{t_1\:=\:0h}$

• $\rm{x_2\:=\:5mm}$

• $\rm{x_1\:=\:0mm}$

$\checkmark\:\rm{\red{\implies\:(2\:-\:0)\:=\:\dfrac{\rho}{10\:k}\:(5^2\:-\:0^2)\:}}$

$\rm{\implies\:2\:=\:\dfrac{\rho}{10\:k}\times{25}\:}$

$\rm{\implies\:\dfrac{\rho}{k}\:=\:2\times{\dfrac{10}{25}}\:}$

$\rm{\implies\:\dfrac{\rho}{k}\:=\:0.8\:}$----(1)

CASE - 2 :-

• Let, $\rm{t_2\:=\:T\:hours}$

• $\rm{t_1\:=\:2h}$

• $\rm{x_2\:=\:20mm}$

• $\rm{x_1\:=\:5mm}$

$\checkmark\:\rm{\red{\implies\:(T\:-\:2)\:=\:\dfrac{\rho}{10\:k}\:(20^2\:-\:5^2)\:}}$

$\rm{\implies\:(T\:-\:2)\:=\:\dfrac{\rho}{10\:k}\:(400\:-\:25)\:}$

$\rm{\implies\:(T\:-\:2)\:=\:\dfrac{\rho}{10\:k}\times{375}\:}$

$\rm{\implies\:(T\:-\:2)\:=\:\dfrac{\rho}{k}\times{\dfrac{375}{10}}\:}$

⚡ Put the value of ‘$\rm\red{\dfrac{\rho}{k}\:=\:0.8\:}$’ in the above equation .

$\rm{\implies\:(T\:-\:2)\:=\:0.8\times{\dfrac{375}{10}}\:}$

$\rm{\implies\:(T\:-\:2)\:=\:0.8\times{37.5}\:}$

$\rm{\implies\:(T\:-\:2)\:=\:30\:}$

$\rm{\implies\:T\:=\:30\:-\:2\:}$

$\rm{\purple{\implies\:T\:=\:28\:hour\:}}$

$\rm\pink{\therefore}$$<font color=baby>$ The time taken for the thickness of ice changes from 5mm to 20mm is 28 hour .