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.