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

Given, `L=3.36xx105Jkg^(-1)`
`K=0.01Wm^(-1)K^(-1)`
`D=0.05M,t=6xx3600=21600s`
`A=0.3xx0.3=0.07m^2,theta_1~~theta_2=45^(@)C`
We know that Q`=(KA(theta_1-theta_2)t)/(d)=mL`
`:.m=(KA(theta_1-theta_2)t)/(Ld)=(0.01xx0.09xx45xx2.16xx10^4)/(3.36xx10^5xx5xx10^(-2))=0.0521kg`
Amount of ice remaining after 6 hours=4-0.0521=3.948kg