Ncert Exercises & Solutions

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 ${Js}^{- 1} m^{- 1} K^{- 1}$ . [Heat of fusion of water = 335 × $10^{3} {Jkg}^{- 1}$ ]

Side of the given cubical icebox, s = 30 cm = 0.3 m

Thickness of the icebox, l = 5.0 cm = 0.05

Time, t = 6 h = 21600 sec

Let m' be the total amount of ice that melts in 6 h.

The total surface area of the box, A $= 6 s^{2} = 6 \times {(0.3)}^{2} = 0.54 m^{2}$

The amount of heat lost by the food:

$Q = \frac{KA ∆ T}{l} t = \frac{0.01 \times 0.54 \times (45) \times 21600}{0.05} = 104976 J$
$∴ Mass of ice melted m' = \frac{Q}{L} = \frac{104976}{335 \times 10^{3}} = 0.313 kg$

Mass of ice left = 4 – 0.313 = 3.687 kg

Hence, the amount of ice remaining after 6 h is 3.687 kg.

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