Thermal Properties of Matter NCERT Solutions Class 11 Physics - Solved Exercise Question 11.19

Question 11.19:
A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. 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 × 103 J kg–1]
Solution:
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 × 103 J kg–1
Let m be the total amount of ice that melts in 6 h.
The amount of heat lost by the food:
θ = KA(T - 0)t / l
Where,
A = Surface area of the box = 6s2 = 6 × (0.3)2 = 0.54 m3
θ = 0.01 X 0.54 X45 X 6 X 60 X 60 / 0.05  =  104976 J
But θ = m'L
∴ m' = θ/L
= 104976/(335 X 103)  =  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.

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