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Exercises · 10.17

Q.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 cm30\ \text{cm} has a thickness of 5.0 cm5.0\ \text{cm}. If 4.0 kg4.0\ \text{kg} of ice is put in the box, estimate the amount of ice remaining after 6 h6\ \text{h}. The outside temperature is 45 ∘C45\ ^\circ\text{C}, and coefficient of thermal conductivity of thermacole is 0.01 J s−1 m−1 K−10.01\ \text{J s}^{-1}\ \text{m}^{-1}\ \text{K}^{-1}. [Heat of fusion of water =335×103 J kg−1= 335 \times 10^{3}\ \text{J kg}^{-1}]

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Heat conducts steadily through all six walls of the icebox and melts the ice at a fixed rate. Over 6 hours, about 0.310.31 kg of ice melts, leaving about 3.693.69 kg remaining.

As long as ice is present, the inside of the box stays at 0∘0^\circC; the outside is at 45∘45^\circC. Heat conducted in through the walls all goes into melting ice (none of it raises the temperature, since the ice holds the inside at its melting point).

Step 1 - Total surface area

The box is a cube of side 30 cm=0.3030\ \text{cm}=0.30 m, with 6 faces:

A=6×(0.30)2=6×0.09=0.54 m2.A = 6\times(0.30)^2 = 6\times0.09 = 0.54\ \text{m}^2.

Step 2 - Rate of heat conduction

Using Fourier's law with wall thickness d=5.0 cm=0.05d=5.0\ \text{cm}=0.05 m, ΔT=45−0=45\Delta T = 45-0=45 K, and k=0.01 J s−1m−1K−1k=0.01\ \text{J s}^{-1}\text{m}^{-1}\text{K}^{-1}:

Qt=kAΔTd=0.01×0.54×450.05=0.2430.05=4.86 W.\frac{Q}{t} = \frac{kA\Delta T}{d} = \frac{0.01\times0.54\times45}{0.05} = \frac{0.243}{0.05} = 4.86\ \text{W}.

Step 3 - Total heat entering in 6 hours …

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