Question

A slab of stone of area $0.36{m^2}$ and thickness $0.1$ is exposed on the lower surface to steam at ${100^ \circ }C$. A block of ice at ${0^ \circ }C$ rests on the upper surface of the slab. In one hour $4.8kg$ of ice melted. The thermal conductivity of slab is: (Given latent heat of fusion of ice=$3.36 \times {10^5}Jk{g^{ - 1}}$)
A) $1.02J/m/s{/^ \circ }C$
B) $1.29J/m/s{/^ \circ }C$
C) $1.24J/m/s{/^ \circ }C$
D) $2.05J/m/s{/^ \circ }C$

Answer 1

Thermal conductivity is measured as the amount of heat flowing through a unit area in unit time with some temperature gradient which is perpendicular to the given area. It does not depend on the nature of the material given. But it varies with temperature, density, or thickness.

Complete step by step solution:
Step I:
Given Latent heat of fusion of ice, $L = 3.36 \times {10^5}Jk{g^{ - 1}}$
Latent heat is the heat required to change the phase of a given mass m and is written as
$dQ = m{L_f}$
Where $dQ$ is the change in heat
$m$ is the mass of the given sample $= 4.8kg$
${L_f}$ is the latent heat of fusion.
Step II:
Fourier’s Law of heat conduction says that the rate of heat transfer through a plane layer is proportional to the temperature gradient across the layer and transfer area of heat.
It is written as:
$\dfrac{{dQ}}{{dT}} = \dfrac{{KA({T_1} - {T_2})}}{x}$---(i)
Where $dQ$ is the rate of heat conduction
$dT$ is the time given $= 1hour = 3600\sec$
$K$ is thermal conductivity of slab $= ?$
$A$ is the area $= 0.36{m^2}$
$x$ is the thickness of the slab $= 0.1m$
${T_1} - {T_2}$ is the difference in temperature $(100 - 0) = 100$
Step III:
Substituting all the values in equation (i),
$\Rightarrow \dfrac{{m{L_f}}}{{dT}} = \dfrac{{KA({T_1} - {T_2})}}{x}$
$\Rightarrow K = \dfrac{{m{L_f}x}}{{\Delta t.A.({T_1} - {T_2})}}$
$\Rightarrow K = \dfrac{{4.8 \times 3.36 \times {{10}^5} \times 0.1}}{{3600 \times 0.36 \times 100}}$
Solving above equation,
$\Rightarrow K = \dfrac{{4.8 \times 3.36}}{{0.36 \times 36}}$
$\Rightarrow K = 1.24J/m/s{/^ \circ }C$

$\therefore$ The thermal conductivity of slab is $1.24J/m/s{/^ \circ }C$. Hence, option (C) is the correct answer.

Note:
Though latent heat describes the energy absorbed or released by any substance with a change in state. But it is of two types: Latent heat of fusion $({L_f})$ and latent heat of vaporization $({L_v})$. So they both are not to be related. The latent heat of fusion is related to the melting of a solid or freezing of a liquid which is discussed above. But latent heat of vaporization is vaporizing a solid or condensing a vapor.