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]

Answer 1

Here,
side length of cubical ice-box = 30cm
area of 6 face of box ( A) = 6 × ( side)²
= 6 × (30)² = 5400 cm²
= 0.54 m²
Thickness of the box ( d) = 5cm = 5 × 10^-2 m
Mass of the ice ( m) = 4 kg
Time ( t) = 6h = 21600 sec
Difference in temperature ( ∆T) = final temperature - initial temperature
= 45°C - 0°C = 45°C
Latent heat of fusion of water ( Lf)= 335 × 10³ j/kg
Coefficient of thermal conductivity ( K) = 0.01 J/s.m.K

Let 'm' is the mass of ice is melted .
Heat supplied by the surrounding= heat taken by ice during melting
KA∆T.t /d = m' Lf
m' = KA∆T.t/Lf.d
= 0.01 × 0.54 × 45 × 21600/335 × 10³×5×10^-2
= 0.313 kg

Hence, mass remains in the box = m-m'
= 4 kg - 0.313 kg
= 3.687 Kg

Answer 2

Here,

side length of cubical ice-box = 30cm

area of 6 face of box ( A) = 6 × ( side)²

= 6 × (30)² = 5400 cm²

= 0.54 m²

Thickness of the box ( d) = 5cm = 5 × 10^-2 m

Mass of the ice ( m) = 4 kg

Time ( t) = 6h = 21600 sec

Difference in temperature ( ∆T) = final temperature - initial temperature

= 45°C - 0°C = 45°C

Latent heat of fusion of water ( Lf)= 335 × 10³ j/kg

Coefficient of thermal conductivity ( K) = 0.01 J/s.m.K

Let 'm' is the mass of ice is melted .

Heat supplied by the surrounding= heat taken by ice during melting

KA∆T.t /d = m' Lf

m' = KA∆T.t/Lf.d

= 0.01 × 0.54 × 45 × 21600/335 × 10³×5×10^-2

= 0.313 kg

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Hence, mass remains in the box = m-m'

= 4 kg - 0.313 kg

= 3.687 Kg