Question

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

Answer 1

Step. 1 Given Data,

Latent heat of fusion of ice, $$L=3.36\times {10}^{5}Jk{g}^{-1}$$

$$m$$ is the mass of the given sample$$=4.8kg$$
$$x$$ is the thickness of the slab$$=0.1m$$
$$A$$is the area$$=0.36m^2$$

A slab of stone is exposed on the lower surface to steam at $T_1$,$${100}^{\circ }C$$

A block of ice rests $T_2=$$$0^{\circ }C$$ rests on the upper surface of the slab
$$dT$$ is the time given$$=1hour=3600\sec$$

Step 2. Formula used,
$$L_f$$is the latent heat of fusion.
$$dQ=m{L}_f$$
$$dQ$$is the change in heat

$$m$$ is the mass of the given sample

$${\displaystyle \frac{dQ}{dT}}={\displaystyle \frac{KA(T_1-T_2)}{x}}$$
$$K$$ is the thermal conductivity of slab

$$A$$ is the area

$dT$ is the change of temperature

$x$ is the thickness

Step 3. Calculating the thermal conductivity
Fourier’s Law of heat conduction is
$\frac{dQ}{dT}=\frac{KA(T_1-T_2)}{x}......\left(i\right)$
$$dT$$ is the time given$$=1hour=3600\sec$$
$$A$$ is the area$$=0.36m^2$$
$$x$$ is the thickness of the slab$$=0.1m$$
$$T_1-T_2$$ is the difference in temperature$$(100-0)=100$$
Substituting all the values in equation $\left(i\right)$,
$$\Rightarrow {\displaystyle \frac{m{L}_f}{dT}}={\displaystyle \frac{KA(T_1-T_2)}{x}}$$
$\Rightarrow K=\frac{m{L}_{f}x}{dT.A.(T_1-T_2)}$
$$\Rightarrow K={\displaystyle \frac{4.8\times 3.36\times {10}^5\times 0.1}{3600\times 0.36\times 100}}$$
Solving the equation,
$$\Rightarrow K={\displaystyle \frac{4.8\times 3.36}{0.36\times 36}}$$
$$\Rightarrow K=1.24J/m/s{/}^{\circ }C$$
$\therefore$ The thermal conductivity of the slab is $$1.24J/m/s{/}^{\circ }C$$.