A slab of stone of area of `0.36 m^(2)` 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 `= 3.63 xx 10^(5) J kg^(-1)`)

A slab of stone of area 3600 cm^(2) and thickness 10 cm is exposed on the lower surface to steam at 100^(@)C . A block of ice at 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 xx 10^(5) j//kg )

A slab of stone area 3600 cm^(2) and thickness 10cm is exposed on the lower surface to steam at 100^(@)C A block of ice at 0^(@)C rest on upper surface of the slab. In one hour 4.8kg of ice is melted. The thermal conductivity of the stone in Js^(-1)m^(-1)k^(-1) is (latent heat of ice =3.36 xx 10^(5) J//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^(@)C .A block of ice at 0^(@)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 xx 10^(4) J kg^(-1) . What is the thermal conductivity of the stone in units of Js^(-1) m^(-1) ""^(@)C^(-1) ?

The lower surface of a slab of stone of face-area 3600cm^(2) and thickness 10cm is exposed to steam at 100^(@)C . A block of ice at 0^(@)C rests on the upper surface of the slab. 4.8g of ice melts in one hour. Calculate the thermal conductivity of the stone. Latent heat of fusion of ice =3.36 xx 10^(5) J kg^(-1)

The lower surface of a slab of stone of face-area 3600cm^(2) and thickness 10cm is exposed to steam at 100^(@)C . A block of ice at 0^(@)C rests on the upper surface of the slab. 4.8g of ice melts in one hour. Calculate the thermal conductivity of the stone. Latent heat of fusion of ice =3.36 xx 10^(5) J kg^(-1)

The lower surface of a slab of stone of face-area 3600cm^(2) and thickness 10cm is exposed to steam at 100^(@)C . A block of ice at 0^(@)C rests on the upper surface of the slab. 4.8g of ice melts in one hour. Calculate the thermal conductivity of the stone. Latent heat of fusion of ice =3.36xx10^(5) J kg^(-1)

The lower surface of a slab of stone of face-area 3600cm^(2) and thickness 10cm is exposed to steam at 100^(@)C . A block of ice at 0^(@)C rests on the upper surface of the slab. 4.8g of ice melts in one hour. Calculate the thermal conductivity of the stone. Latent heat of fusion of ice =3.36xx10^(5) J kg^(-1)

A block of ice at 0^(@)C rest on the upper surface of the slab of stone of area 3600cm^(2) and thickness of 10cm . The slab is exposed on the lower surface to steam at 100^(@)C . If 4800g of ice is melted in one hour, then calculate the thermal conductivity of stone. (Given the latent heat of fusion of ice =80cal//g )

A stone of 0.36 m^(2) cross -sectional area and 0.1 m thickness has its lower surface in contact with water at 100^(@)C and upper surface in contact with ice at 0^(@)C . If 4.8 kg ice melting in 1 hour, then what is the thermal conductivity of stone ? Latent heat of fusion of ice is 3.36xx10^(5)" J/kg" .