A ‘thermacole’ icebox is a cheap and an efficient method for stor… - Vidyadip

Question

A ‘thermacole’ icebox is a cheap and an efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30cm has a thickness of 5.0cm. If 4.0kg of ice is put in the box, estimate the amount of ice remaining after 6h. The outside temperature is 45°C, and co-efficient of thermal conductivity of thermacole is 0.01 J s^–1 m^–1 K^–1. [Heat of fusion of water = 335 × 103J kg^–1].

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Answer

Side of the given cubical ice box, s = 30cm = 0.3m
Thickness of the ice box, l = 5.0cm = 0.05m
Mass of ice kept in the ice box, m = 4kg
Time gap, t = 6h = 6 × 60 × 60s
Outside temperature, T = 45°C
Coefficient of thermal conductivity of thermacole, K = 0.01 J s^–1m^–1K^–1
Heat of fusion of water, L = 335 × 10³J kg^–1
Let m^’ be the total amount of ice that melts in 6h.
The amount of heat lost by the food:
$\theta=\text{KA}(\text{T}-0)\frac{\text{t}}{1}$
Where,
A = Surface area of the box = 6s² = 6 × (0.3)² = 0.54m³
$\theta=0.01\times0.54\times45\times6\times60\times\frac{60}{0.05}=104976\text{J}$
But $\theta=\text{m}'\text{L}$
$\therefore\text{m}'=\frac{\theta}{\text{L}}$
$=\frac{104976}{(335\times10^3)}=0.313\text{kg}$
Mass of ice left = 4 – 0.313 = 3.687kg
Hence, the amount of ice remaining after 6 h is 3.687kg.

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