Question

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 the slab is:
(Given latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹)

विकल्प

• 1.24 J/m°C

• 1.29 J/m°C

• 2.05 J/m°C

• 1.02 J/m°C

Answer 1

1.24 J/m°C

Explanation:

Rate of heat given by steam = Rate of heat taken by the 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"/"dt" = ("KA"(100 - 0))/1 = "m""dL"/"dt"`,

`("K" xx 100 xx 0.36)/0.1 = (4.8 xx 3.36 xx 10^5)/(60 xx 60)`

K = 1.24 J/m/s/°C