question_answer

A slab of stone of area \[0.36 {{m}^{2}}\]and thickness 0.1 m is exposed on the lower surface to steam at\[100{}^\circ C\]. A block of ice at \[0{}^\circ C\] 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 effusion of ice \[=3.36\times 105 J\]\[{{\operatorname{kg}}^{-1}}\].):  

A) \[1.24J/m{}^\circ C\]                

B) \[1.29J/m{}^\circ C\]

C) \[2.05J/m{}^\circ C\]    

D)        \[1.02J/m{}^\circ C\]

Correct Answer: A

Solution :

[a]  \[\frac{dQ}{dt}=\frac{KA\left( 100-0 \right)}{1}=m\frac{dL}{dt},\]  \[\frac{K\times 100\times 0.36}{0.1}=\frac{4.8\times 3.36\times {{10}^{5}}}{60\times 60}\] \[=1.24J/m/s/{}^\circ C\]