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Mechanical Properties Of Fluids

Question
CBSEENPH11020043

A ‘thermocole’ 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 thermocole is 0.01 J s–1 m–1 K–1. [Heat of fusion of water = 335 × 103 J kg–1]

Solution

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
Outside temperature, T = 45°C
Coefficient of thermal conductivity of thermocole, 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, 

straight theta space equals space fraction numerator KA left parenthesis straight T minus 0 right parenthesis straight t over denominator straight l end fraction

where comma space

straight A space equals space surface space area space of space box space equals space 6 space straight s squared space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 6 cross times left parenthesis 0.3 right parenthesis squared space

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.54 space straight m cubed space

straight theta space equals space fraction numerator 0.01 space cross times space 0.54 cross times space 45 space cross times space 6 space cross times space 60 space cross times space 60 over denominator 0.05 end fraction
space space space equals space 104976 space straight J

But comma space

space space space space space space space space straight theta space equals space straight m apostrophe straight L

therefore space straight m apostrophe space equals space straight theta over straight L
space space space space space space space space space space equals fraction numerator 104976 over denominator 335 cross times 10 cubed end fraction space equals space 0.313 space kg

Mass space of space ice space left space equals space 4 space minus space 0.313 space equals space 3.687 space kg

That space is comma space

Amount space of space ice space remaining space after space 6 space straight h space is space 3.687 space kg. space