Estimate the rate at which ice would melt in a wooden box 2.5 mm thick and of inside measurement 100 cm x 60 cm x 40 cm, assumming that external temperature is `32 .^(@)C` and coefficient of thermal conductivity of wood is `0.168 Wm^(-1) K^(-1)`. Given `L=80 cal//g`.

Answer

`[1.587 g s^(-1)]`

Explanation

Here, `Delta x=2.5 cm=2.5xx10^(-2)m`,
`A=(100xx60+60xx40+40xx100)2 cm^(2)`,
`=24800 cm^(2) =2.48 m^(2)`,
`Delta T=32-0=32K`,
`K=0.168 Wm^(-1) K^(-1)=(0.168)/(4.2)cal m^(-1) K^(-1)`
If m is the mass (in gram) of ice melted per second, then `(dQ)/(dt)=mL KA(Delta T)/(Delta x)`
or `m=(KA Delta T)/(L Delta x)=((0.168//4.2)xx2.48xx32)/(80xx2.5xx20^(-2)) =1.587 g`