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 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 $3.36×{10}^{5}Jk{g}^{-1}$)

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Solution

## Step. 1 Given Data,Latent heat of fusion of ice, $L=3.36×{10}^{5}Jk{g}^{-1}$$m$ is the mass of the given sample$=4.8kg$$x$ is the thickness of the slab$=0.1m$$A$is the area$=0.36{m}^{2}$A slab of stone is exposed on the lower surface to steam at ${T}_{1}$,${100}^{\circ }C$A block of ice rests ${T}_{2}=$${0}^{\circ }C$ rests on the upper surface of the slab$dT$ is the time given$=1hour=3600\mathrm{sec}$Step 2. Formula used,${L}_{f}$is the latent heat of fusion.$dQ=m{L}_{f}$$dQ$is the change in heat$m$ is the mass of the given sample$\frac{dQ}{dT}=\frac{KA\left({T}_{1}-{T}_{2}\right)}{x}$$K$ is the thermal conductivity of slab$A$ is the area$dT$ is the change of temperature$x$ is the thicknessStep 3. Calculating the thermal conductivityFourier’s Law of heat conduction is$\frac{dQ}{dT}=\frac{KA\left({T}_{1}-{T}_{2}\right)}{x}......\left(i\right)$$dT$ is the time given$=1hour=3600\mathrm{sec}$$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$\left(100-0\right)=100$Substituting all the values in equation $\left(i\right)$,$⇒\frac{m{L}_{f}}{dT}=\frac{KA\left({T}_{1}-{T}_{2}\right)}{x}$$⇒K=\frac{m{L}_{f}x}{dT.A.\left({T}_{1}-{T}_{2}\right)}$$⇒K=\frac{4.8×3.36×{10}^{5}×0.1}{3600×0.36×100}$Solving the equation,$⇒K=\frac{4.8×3.36}{0.36×36}$$⇒K=1.24J/m/s{/}^{\circ }C$$\therefore$ The thermal conductivity of the slab is $1.24J/m/s{/}^{\circ }C$.

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