NEETPhysicsProperties of MatterMCQ+4 / −1
A slab of stone of area 0.36 m2 and thickness 0.1 m is exposed on the lower surface to steam at 100oC. A block of ice at 0oC 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 kg1)
(Given latent heat of fusion of ice = 3.36 105 J kg1)
- A1.24 J/m/s/oC
- B1.29 J/m/s/oC
- C2.05 J/m/s/oC
- D1.02 J/m/s/oC
View written solutionFree
Correct answer: 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
,
K =1.24 J/m/s/°C
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