11.17 A ‘thermacole’ icebox is a cheap and an 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]
11.17 A ‘thermacole’ icebox is a cheap and an 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]
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1 Answer
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11.17 Size of the sides of cubical ice box, s = 30 cm =0.3 m
Thickness of the icebox, l = 5 cm = 0.05 m
Mass of ice kept in the box, m = 4 kg
Time gap, t = 6 h = 6 s
Outside temperature, T = 45 °C
Coefficient of thermal conductivity of thermocole, K = 0.01 J
Heat of fusion of water, L = 335 J
Let m’ be the mass of the ice melts in 6 h
The amount of heat lost by the food: = , where
A = Surface area of the box = 6 = 6 = 0.54
= = 104976 J
We also know so m’ = 104976/ (335 = 0.313 kg
Hence the amount of ice remains
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Heat Released by block = Heat gain by large Ice block
5 × 0.39 × 500 = mice × 335
= 2.91 kg
Given
6a2 = 24
a2 = 4
a = 2m
This is a multiple choice answer as classified in NCERT Exemplar
(b), (c), (d) When the hot milk in the table is transferred to the surroundings by conduction, convection and radiation.
According to newton's law of cooling temperature of the milk falls of exponentially. Heat also will be transferred from surroundings to the milk but will be lesser than that of transferred from milk to surroundings. So option b, c, d satisfy.
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