A cubical thermocol ice box of side length `30cm` has thickess of `5.0cm` If `4.0kg` of ice is put in the box estimate the amount of ice remaining after `6hr` The outside temperature is `45^(@)C` and co-efficient of the thermal conductivity of the thermocol is `0.01 J s^(-1) m^(-1) K^(1)` (Latent Heat of fusion of water `=335 xx 10^(3) Jkg^(-1))` .

Explanation

Here, length of each side I = 30 cm = 0.3 m Thickness of each side, ` Delta x = 5 cm = 0.05 m`
Total surface area through which heat enters into the box, `A = 6 l^2 = 6 xx 0.3 xx0.3 = 0.54 m^2`
Temp diff, `Delta T =45 -0 = 45^@C, K = 0.01 J s^(-1) m^(9-1) ^@C^(-1)`
time, `Delta t = 6hrs = 6xx60 xx 60s`
Latent heat of fusion `L = 335 xx 10^3 j//kg`
Let m be the mass of ice melted in this time `:. Delta Q = mL = KA ((Delta T)/(Deltax)) Delta t`
`m =KA ((DeltaT)/(Delta x)) (Deltat)/(L) = 0.01 xx 0.54 xx(45)/(0.05) xx(6xx60xx60)/(335 xx 10^3) = 0.313 kg`
`:.` Mass of ice left `= 4 - 0.313 = 3.687 kg`