A cubical thermocol box, full of …

The quantity of heat Q transferred into the box  from its one face 

Q = {tex}KA(T_h–T_c)t\over d{/tex}  

where K is thermal conductivity of thermocol.

=  {tex}0.01×2500×10^{–4}×40º×5×60×60\over 4×10^{–2}{/tex} = 45000 J 

As the box has six faces, total heat passing into the box Q = 45000 × 6 J = 270000 J 

The mass m of ice melted= {tex} Q\over L {/tex} = {tex}270000\over 335{/tex} = 805.97 g

The quantity of heat transferred into the box through its one face can be obtained by equation :

Q = {tex}KA (T_h –T_c ) t\over d {/tex} = 

where K is thermal conductivity of thermocol.

A is area of one face of cube. 

{tex}T_h \ is \ outside \ temperature{/tex}

{tex}T_c \ is \ inside \ temperature{/tex}

t is time and d is thickness.

So 

=  {tex} 0.01 × 2500 × 10^{–4} × 40º  × 5 ×  60 × 60 \over 4 × 10^{–2  }{/tex} = 45000 J 

Since the box has six faces, total heat passing into the box Q = 45000 × 6 J

The mass of ice melted  m, can be obtained by dividing  Q  by  L  :

m =  Q/L ={tex}45000 × 6 \over 335{/tex} = 805.97 g