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 the 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]

Asked by Abhisek | 1 year ago |  183

1 Answer

Solution :-

Side of the cubical icebox, s =30 cm=3 x 10-2 m
Thickness of the icebox, L =5.0 cm=0.05 m
Mass of ice kept in the icebox, m=4 kg

Time, t=6 h=6×60×60 = 21600
Outside temperature, T1= 45°C

Temperature of the icebox = 0°C

Temperature difference = T1 – T2 =45°C – 0°C

Surface area of the icebox = 6 x (0.30)2= 0.54
Coefficient of thermal conductivity of thermacole, K=0.01 Js−1m−1k−1
Heat of fusion of water, L= 335×10 3 Jkg −1

Total heat entering the icebox in 6 hours is

Q = \( \dfrac{KA(T_1−T_2)t}{L}\)

\(\dfrac{(0.01 Js^{-1}m^{-1}C^{-1} \times 0.54 m^2 \times 450 C \times 21600 s)}{0.05 m}\)

= 1.05 x 105 J

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

But Q= mL
Therefore, m = Q/L
=\( \dfrac{1.05 \times 10^5}{(335×10^3)}\)

=0.313 kg

Amount of ice remaining after 6 h

= 4–0.313

=3.687 kg

Answered by Pragya Singh | 1 year ago

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