
Ice starts freezing in a lake with water at ${{0}^{{}^\circ }}C$. If the time taken for 1 cm of ice to be formed is 12 min, the time taken for the thickness of the ice to change from 1 cm to 2 cm will be
A. 12 min
B. less than 12 min
C. more than 12 min but less than 24 min
D. more than 24 min
Answer
164.1k+ views
Hint: To solve this question, first we made a relation between the time takes to melt a certain layer with the length of ice layer by calculating the amount of heat required to melt a very small layer of ice and then integrate it to get a desirable answer.
Formula used:
The formula of rate of flow of heat is,
$\dfrac{dQ}{dt}=\dfrac{KA(\Delta T)}{dx}$
Where, $K$ is the thermal conductivity, $A$ is the area of cross-section and $\Delta T$ is the change in temperature.
Complete step by step solution:
Consider x be the length of ice that has already formed and it takes dt time to form dx layer of ice over a surface area of A with thermal conductivity of ice being K. Let the mass of ice be m, its density $\rho $ and then the volume of ice will be Ax. So the heat transferred be,
$\dfrac{dQ}{dt}=\dfrac{KA(\Delta T)}{dx} \\ $
It can also be written as,
$\dfrac{d(mL)}{dt}=\dfrac{KA(\Delta T)}{dx} \\ $
And, $\dfrac{d(\rho AxL)}{dt}=\dfrac{KA(\Delta T)}{dx} \\ $
$dt=\dfrac{\rho L}{K\Delta T}xdx \\ $
Where L is the latent heat of ice fusion.
Equation can be write as $dt=Cxdx$
Where $C=\dfrac{\rho L}{K\Delta T}$
Now we integrate the equation in limits $(0\to t)$ for time $(0\to l)$ for the length of ice formed
We get $\int_{0}^{t}{dt=\int_{0}^{l}{Cxdx}}$
Integrating the above equation, we get
$t=C\dfrac{{{l}^{2}}}{2}$
$\therefore \dfrac{t}{{{l}^{2}}}=\text{constant}$
Now we can write the above equation for two different layers of ice formed as :
$\dfrac{{{t}_{1}}}{{{l}_{1}}^{2}}=\dfrac{{{t}_{2}}}{{{l}_{2}}^{2}}$
Now we put the values given in the question as
$\dfrac{12}{{{1}^{2}}}=\dfrac{{{t}_{2}}}{{{2}^{2}}}$
Then ${{t}_{2}}=48$ minutes
Thus 48 min is the time taken to form 2 cm of ice and 12 min is the time to form 1 cm of ice from the start. Then to form the second 1 cm of ice, the time taken is,
${{t}_{2}}-{{t}_{1}}= 48-12= 36$ minutes
Thus, the time taken for the thickness of the ice to change from 1 cm to 2 cm will be more than 24 minutes.
Thus, option D is the correct answer.
Note: Whenever we solve these types of questions, we use integration to solve it. With the help of integration, we can solve it in less time and also get the correct answer but we must know the basic rules of integration to solve it.
Formula used:
The formula of rate of flow of heat is,
$\dfrac{dQ}{dt}=\dfrac{KA(\Delta T)}{dx}$
Where, $K$ is the thermal conductivity, $A$ is the area of cross-section and $\Delta T$ is the change in temperature.
Complete step by step solution:
Consider x be the length of ice that has already formed and it takes dt time to form dx layer of ice over a surface area of A with thermal conductivity of ice being K. Let the mass of ice be m, its density $\rho $ and then the volume of ice will be Ax. So the heat transferred be,
$\dfrac{dQ}{dt}=\dfrac{KA(\Delta T)}{dx} \\ $
It can also be written as,
$\dfrac{d(mL)}{dt}=\dfrac{KA(\Delta T)}{dx} \\ $
And, $\dfrac{d(\rho AxL)}{dt}=\dfrac{KA(\Delta T)}{dx} \\ $
$dt=\dfrac{\rho L}{K\Delta T}xdx \\ $
Where L is the latent heat of ice fusion.
Equation can be write as $dt=Cxdx$
Where $C=\dfrac{\rho L}{K\Delta T}$
Now we integrate the equation in limits $(0\to t)$ for time $(0\to l)$ for the length of ice formed
We get $\int_{0}^{t}{dt=\int_{0}^{l}{Cxdx}}$
Integrating the above equation, we get
$t=C\dfrac{{{l}^{2}}}{2}$
$\therefore \dfrac{t}{{{l}^{2}}}=\text{constant}$
Now we can write the above equation for two different layers of ice formed as :
$\dfrac{{{t}_{1}}}{{{l}_{1}}^{2}}=\dfrac{{{t}_{2}}}{{{l}_{2}}^{2}}$
Now we put the values given in the question as
$\dfrac{12}{{{1}^{2}}}=\dfrac{{{t}_{2}}}{{{2}^{2}}}$
Then ${{t}_{2}}=48$ minutes
Thus 48 min is the time taken to form 2 cm of ice and 12 min is the time to form 1 cm of ice from the start. Then to form the second 1 cm of ice, the time taken is,
${{t}_{2}}-{{t}_{1}}= 48-12= 36$ minutes
Thus, the time taken for the thickness of the ice to change from 1 cm to 2 cm will be more than 24 minutes.
Thus, option D is the correct answer.
Note: Whenever we solve these types of questions, we use integration to solve it. With the help of integration, we can solve it in less time and also get the correct answer but we must know the basic rules of integration to solve it.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
