A slab of stone of area 0-36 m2 and thickness 0-1 m is exposed on the lower surface to steam at 100-C- A block of ice at 0-C rests on the upper surface of the slab- In one hour 4-8 kg of ice is melted- The thermal conductivity of slab is Given- latent heat of fusion of ice - 3-36 - 105 J kg-1

Given A=0.36 m^2, L=0.1 m, T_1=0^∘C, T_2=100^∘C, m=4.8 kg, t=1 hour, H= 3.36 × 10^5 J kg^–1 Heat transfer through slab in one hour= Heat required to me ...

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Standard VIII

Physics

Change of State

A slab of sto...

Question

A slab of stone of area $0.36 m^{2}$ and thickness $0.1 m$ 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 $4.8 kg$ of ice is melted. The thermal conductivity of slab is (Given: latent heat of fusion of ice $= 3.36 \times 10^{5} {J kg}^{- 1}$ )

1.29 J/m/s /^{\circ} C

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2.05 J/m/s /^{\circ} C

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1.02 J/m/s /^{\circ} C

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1.24 J/m/s /^{\circ} C

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Solution

The correct option is D $1.24 J/m/s /^{\circ} C$
Given-
$A = 0.36 m^{2}, L = 0.1 m,$
$T_{1} = 0^{\circ} C, T_{2} = 100^{\circ} C,$
$m = 4.8 kg, t = 1 hour, H = 3.36 \times 10^{5} {J kg}^{- 1}$

Heat transfer through slab in one hour $=$ Heat required to melt ice

$q = \frac{K A (Δ T) t}{L} = m L$
$\frac{K \times 0.36 (100 - 0) \times 3600}{0.1} = 4.8 \times 3.36 \times 10^{5}$
$K = 1.24 J/m/s /^{\circ} C$ .
Hence, option $(B)$ is correct.

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A slab of stone of area 0.36m2 and thickness 0.1m is exposed on the lower surface to steam at 100∘C. A block of ice at 0∘C rests on the upper surface of the slab. In one hour 4.8kg of ice is melted. The thermal conductivity of slab is (Given: latent heat of fusion of ice =3.36×105 J kg–1)

The correct option is D 1.24J/m/s/∘CGiven- A=0.36m2, L=0.1m, T1=0∘C, T2=100∘C, m=4.8kg,t=1hour,H=3.36×105 J kg–1 Heat transfer through slab in one hour= Heat required to melt ice q=KA(ΔT)tL=mL K×0.36(100−0)×36000.1=4.8×3.36×105 K=1.24J/m/s/∘C. Hence, option (B) is correct.