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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
VerifiedVerified
161.1k+ 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.