A slab of stone of area 0.36 m2 and thickness 0.1 m is exposed on the lower surface to steam at 100 . A block of ice at 0 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)
- 1.24 J/m/s/ C
- 1.29 J/m/s/ C
- 2.05 J/m/s/ C
- 1.02 J/m/s/ C
Answer
(1) 1.24 J/m/s/ C
Solution
Heat conducted in 1 h = heat used to melt the ice:
Q = mL = 4.8 3.36 105 = 1.6128 106 J
Also Q = KA(T1 - T2)td
K = QdA(T1 - T2)t
= 1.6128 106 0.10.36 100 3600
= 161280129600 1.24 J m-1 s-1 C-1
Answer: (1)