A thermocole cubical icebox of side 30 cm has a thickness of 5.0 cm if 4.0 kg of ice are put ini the box, estimate the amount of ice remaining after 6 h. The outside temperature is `45^@C` and coefficient of thermal conductivity of thermocole`=0.01 J//kg`.

Explanation

Side of the given cubical ice box, s = 30 cm = 0.3 m
Thickness of the ice box, l = 5.0 cm = 0.05 m
Mass of ice kept in the ice box, m = 4 kg
Time gap, t = 6 h = 6 × 60 × 60 s
Outside temperature, `T = 45^(@)C`
Coefficient of thermal conductivity of thermacole, `K = 0.01 J s^(–1) m^(–1) K^(–1)`
Heat of fusion of water,` L = 335 × 10^(3) J kg^(–1)`
Let m’ be the total amount of ice that melts in 6 h.
The amount of heat lost by the food:
`theta=(KA(T-0)t)/(t)`
Where,
A = Surface area of the box `= 6s^(2) = 6 × (0.3)^(2) = 0.54 m^(3)`
`theta=(0.01xx0.54xx(45)xx6xx60xx60)/(0.05)=104976 J`
But `theta=m'L`
`:.m'=(theta)/(L)`
`=(104976)/(335xx10^(3))=0.313 kg`
Mass of ice left = 4 – 0.313 = 3.687 kg
Hence, the amount of ice remaining after 6 h is 3.687 kg.