A slab of stone of area 0.36 m² and thickness 0.1 m

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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 .) :

1. 1.24 J/m/s/°C
2. 1.29 J/m/s/°C
3. 2.05 J/m/s/°C
4. 1.02 J/m/s/°C

Correct Option: A

Rate of heat given by steam = Rate of heat taken by ice
where K = Thermal conductivity of the slab
m = Mass of the ice
L = Latent heat of melting/fusion
A = Area of the slab

	DQ	=	KA(100 - 0)	= m	dL	,
	dt		1		dt

	K × 100 × 0.36	=	4.8 × 3.36 × 10⁵
	0.1		60 × 60

K = 1.24 j/m/s/°C