A slab of stone are 3600 cm2 and thickness 10 cm is exposed on the lower surface to steam at 100-C- A block of ice of 0-C rests on upper surface of the slab- In one hour 4-8 kg of ice is melted- The thermal conductivity of the stone in Js-1 m-1 k-1 is Latent heat of ice - 3-36- 105J-kg

A slab of stone of area $0.34 m^{2}$ and thickness 10 cm is exposed on the lower face to steam at $100^{\circ} C$ . A block of ice at $0^{\circ} C$ rests on the upper face of the slab. In one hour, 3.6 kg of ice is melted. Assume that the heat loss from the sides is negligible. The latent heat of fusion of ice is $3.4 \times 10^{4} J k g^{- 1}$ . What is the thermal conductivity of the stone in units of $J s^{- 1} m^{- 1} C^{- 1}$ ?

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A slab of stone are 3600 cm2 and thickness 10 cm is exposed on the lower surface to steam at 100∘C. A block of ice of 0∘C rests on upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of the stone in Js−1m−1k−1 is (Latent heat of ice =3.36×105J/kg)

The correct option is D 1.24Qt=KA(Q1Q2)l=48×3.36×10560×60=K×3600×10−4×10010×10−2=1.24

The correct option is D $1.24$
$\frac{Q}{t} = \frac{K A (Q_{1} Q_{2})}{l} = \frac{48 \times 3.36 \times 10^{5}}{60 \times 60} = \frac{K \times 3600 \times 10^{- 4} \times 100}{10 \times 10^{- 2}} = 1.24$