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]

Question

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]

Philoid · Accepted Answer

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.