# 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 - Physics

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 coefficient 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 1

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: theta = (KA(T-0)t)/l

Where

A = Surface area of the box = 6s2 = 6 × (0.3)2 = 0.54 m3

theta = (0.01xx0.54xx(45)xx6xx60xx60)/0.05 = 104976 J

But theta = m'L

:. m' =  theta/L

= 104976/(335xx10^3) = 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.

#### Solution 2

Each side of the cubical box (having 6 faces) is 30 cm = 0.30 m. Therefore, the total surface area’ of the icebox exposed to outside air is A = 6 x (0.30 m)2 = 0.54 m2. The thickness of the icebox is d = 5.0 cm = 0.05 m, time of exposure t = 6h = 6 x 3600 s and temperature difference T1 – T2 = 45°C – 0°C = 45°C.

∴Total heat entering the icebox in 6 h is given by

Q = (KA(T1 - T2)t)/d

= (0.01 Js^(-1) m^(-1) C6(-1)xx0.54 m^2 xx 45 ^@C xx (6xx 3600 s))/"0.05m"

= 1.05 xx10^5 J

Suppose a mass m of ice melts with this heat. Then Q = mL, where L is tatent heat of fusion of water thus

1.05 xx 10^5 J = m(335xx10^3) "JKg"^(-1)

or m = (1.05 xx10^5 J)/(335 xx 10^3 J kg^(-1)) = 0.313 kg

The initial mass of ice in the box is 4.0 kg. Therefore the ice remaining in the box after 6 h is  = (4.0 - 0.313) kg

= 3.687 kg

#### Solution 3

Each side of the cubical box (having 6 faces) is 30 cm = 0.30 m. Therefore, the total surface area’ of the icebox exposed to outside air is A = 6 x (0.30 m)2 = 0.54 m2. The thickness of the icebox is d = 5.0 cm = 0.05 m, time of exposure t = 6h = 6 x 3600 s and temperature difference T1 – T2 = 45°C – 0°C = 45°C.

∴Total heat entering the icebox in 6 h is given by

Q = (KA(T1 - T2)t)/d

= (0.01 Js^(-1) m^(-1) C6(-1)xx0.54 m^2 xx 45 ^@C xx (6xx 3600 s))/"0.05m"

= 1.05 xx10^5 J

Suppose a mass m of ice melts with this heat. Then Q = mL, where L is tatent heat of fusion of water thus

1.05 xx 10^5 J = m(335xx10^3) "JKg"^(-1)

or m = (1.05 xx10^5 J)/(335 xx 10^3 J kg^(-1)) = 0.313 kg

The initial mass of ice in the box is 4.0 kg. Therefore the ice remaining in the box after 6 h is  = (4.0 - 0.313) kg

= 3.687 kg

Is there an error in this question or solution?

#### APPEARS IN

NCERT Class 11 Physics Textbook
Chapter 11 Thermal Properties of Matter
Q 19 | Page 296