A thermocol ice box is cheap and efficient method for storing small quantities of cooked food in a summer particular. a cubical ice box of side side 30 cm has a thickness of 5 cm if 4 kg of ice is put in the box estimate the amount of ice remaining after 6h . the outside temperature is 45°C and coefficient of thermal conductivity of thermocol is 0.01j(smk)^-1 heat of fusion of water is 355 *10³jkg^-1

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 thermoCol, K = 0.01 J s–1 m–1 K–1

Heat of fusion of water, L = 335 × 103 J kg–1

Let m’ = total amount of ice that melts in 6 h.

The amount of heat lost by the food:

θ = (KA (T1 – T2)/t)/d  

Where,

A = Surface area of the box = 6s2 = 6 × (0.3)2 = 0.54 m3

θ = (0.01 x 0.54 x (45) x 6 x 60 x 60)/ (0.05)

=104976 J

But θ = m’L

Or m’ = (θ /L)

= (104976)/ (335 x 103)

=0.313Kg

Mass of ice left = (4 – 0.313) = 3.687 kg

Hence, the amount of ice remaining after 6 h is 3.687 kg.