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]
Given,
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 = 21600 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 = 6×s×2 = 6×0.3×2 = 0.54 m3
∴ θ = [(0.01 Js–1m–1K–1)×(0.54m3)×(45°C)×(21600s)]/(0.05 m)
⇒ θ = 104976 J
But θ = m'L
∴ m’ = θ/L = 104976/(335 × 103 J kg-1)
⇒ m’ = 0.313 g
Mass of ice left = 4 – 0.313 = 3.687 kg
Hence, the amount of ice remaining after 6 h is 3.687 kg.
NOTE: Heat of fusion is the amount of heat supplied to a specific quantity of the substance to change its state from a solid to a liquid at constant pressure.