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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 45oC, and co-efficient of thermal conductivity of thermacole is 0.01Js1m1K1. [Heat of fusion of water =335×103Jkg1]

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Solution

Side of the given cubical ice box, s=30cm=0.3m
Thickness of the ice box, l=5.0cm=0.05m
Mass of ice kept in the ice box, m=4kg
Time gap, t=6h=6×60×60s
Outside temperature, T=45oC
Coefficient of thermal conductivity of thermacole, K=0.01Js1m1k1
Heat of fusion of water, L=335×103Jkg1
Let m be the total amount of ice that melts in 6 h.
The amount of heat lost by the food:
θ=KA(T0)t/l
Where,
A= Surface area of the box =6s2=6×(0.3)2=0.54m3
θ=0.01×0.54×45×6×60/0.05=104976J
But θ=mL
m=θ/L
=104976/(335×103)=0.313kg
Mass of ice left =40.313=3.687kg
Hence, the amount of ice remaining after 6 h is 3.687 kg.

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