A cubical ice box of thermocol has each side 30 cm and thickness 5 cm , 4 kg of ice is put in the box , if outside temperature is `45^(@)C` and coefficient of thermal conductivity is `0.01 J s^(-1) m^(-1) K^(-1)` . Calculate the mass of ice left after 6 hrs . Take latent heat of fusion of ice as `335 xx 10^(3) JK^(-1)`

Explanation

`A = 1 xx b = 30 xx 30 cm^(2)`
`A - = 900 cm^(2) = 900 xx 10^(-4) m^(2)` ,
d = 5 cm = `5 xx 10^(-2) m , m = 4 kg " " K = 0.01 Wm^(-1) k^(-1) , M_("ice")` = ?
`L = 335 xxx 10^(3) J kg^(-1) theta_(1) = 45^(@) C , theta_(2) = 0^(@) C`
`t = 6 xx 3600 = 19600 = 1.96 xx 10^(4) s`
We know that `mL = (KA (theta_(1) - theta_(2))t)/(l)`
i.e. ` m = (KA (theta_(1) - theta_(2)) t)/(dL)`
i.e. ` m = (0.1 xx 9 xx 10^(-2) xx (45-0) xx 1.96 xx 10^(4))/(5 xx 10^(-2) xx 3.35 xx 10^(5))`
i.e. m `= (793 . 8 xx 10^(-4 + 4))/(16.75 xx 10^(3)) = (793.8)/(16.7 xx 10^(3))` kg .
i.e. `m = 0.0475 `kg
Hence mass of ice left over after 6 hr = 4 kg `- 0.0475 ` kg = 3.9525 kg .