
One end of copper rod of length 1.0 m and area of cross- section is ${{10}^{-2}}{{m}^{2}}$ is immersed in boiling water and the other end in ice. If the coefficient of thermal conductivity of copper is $92\,cal/m{{s}^{-1}}$, and the latent heat of ice is $8\times {{10}^{4}}\,cal/kg$, then the amount of ice which will melt in one minute is
A. $9.2\times {{10}^{-3}}kg$
B. $8\times {{10}^{-3}}kg$
C. $6.9\times {{10}^{-2}}kg$
D. $5.4\times {{10}^{-3}}kg$
Answer
164.4k+ views
Hint:In this question, we have to find the amount of ice melt. First we find heat generated per minute by putting the values in the formula of heat conduction and then by using latent heat, we find the amount of ice melt.
Formula used:
We used the heat conduction formula
H = $KA\dfrac{dt}{dx}\times t$
Where K is the thermal conductivity, A is the area of cross section, T is the time and $\dfrac{dt}{dx}$ is the temperature difference.
Complete step by step solution:
Given length of copper rod, $dx=1m$
Area of cross- section (A) = ${{10}^{-2}}{{m}^{2}}$
Coefficient of thermal conductivity of copper, K = $92cal/m{{s}^{-1}}$
Latent heat of ice, L = $8\times {{10}^{4}}cal/kg$
Here, One end of copper rod of length 1.0 m and area of cross- section ${{10}^{-2}}{{m}^{2}}$ is immersed in boiling water and the other end in ice.
So, the heat generated per minute,
H = $KA\dfrac{dt}{dx}\times t$
Substituting the values in the above equation, we get
$H = 92\times {{10}^{-2}}\times \dfrac{(100-0)}{1}\times 60\,cal \\ $
$\Rightarrow H = 92\times 60\,cal \\ $
$\Rightarrow H=5520\,cal$
Heat needed to melt the ice of mass,
$m = m\times 8\times {{10}^{4}}=5520$
$\Rightarrow m = \dfrac{5520}{8}\times {{10}^{4}}$
$\therefore m=6.9\times {{10}^{-2}}kg$
The quantity of ice which will melt in one minute is $6.9\times {{10}^{-2}}kg$.
Thus, option C is the correct answer.
Note: Thermal conductivity of a material is the measure of the ability of material to conduct the heat. Heat transfer occurs at a slow pace in the materials of low thermal conductivity and at a higher pace in the materials of higher conductivity.
Formula used:
We used the heat conduction formula
H = $KA\dfrac{dt}{dx}\times t$
Where K is the thermal conductivity, A is the area of cross section, T is the time and $\dfrac{dt}{dx}$ is the temperature difference.
Complete step by step solution:
Given length of copper rod, $dx=1m$
Area of cross- section (A) = ${{10}^{-2}}{{m}^{2}}$
Coefficient of thermal conductivity of copper, K = $92cal/m{{s}^{-1}}$
Latent heat of ice, L = $8\times {{10}^{4}}cal/kg$
Here, One end of copper rod of length 1.0 m and area of cross- section ${{10}^{-2}}{{m}^{2}}$ is immersed in boiling water and the other end in ice.
So, the heat generated per minute,
H = $KA\dfrac{dt}{dx}\times t$
Substituting the values in the above equation, we get
$H = 92\times {{10}^{-2}}\times \dfrac{(100-0)}{1}\times 60\,cal \\ $
$\Rightarrow H = 92\times 60\,cal \\ $
$\Rightarrow H=5520\,cal$
Heat needed to melt the ice of mass,
$m = m\times 8\times {{10}^{4}}=5520$
$\Rightarrow m = \dfrac{5520}{8}\times {{10}^{4}}$
$\therefore m=6.9\times {{10}^{-2}}kg$
The quantity of ice which will melt in one minute is $6.9\times {{10}^{-2}}kg$.
Thus, option C is the correct answer.
Note: Thermal conductivity of a material is the measure of the ability of material to conduct the heat. Heat transfer occurs at a slow pace in the materials of low thermal conductivity and at a higher pace in the materials of higher conductivity.
Recently Updated Pages
How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

Environmental Chemistry Chapter for JEE Main Chemistry

Uniform Acceleration - Definition, Equation, Examples, and FAQs

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

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
