A 'thermacole' icebox is a cheap and an efficient method for stor… - Vidyadip

Question

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 $45^{\circ} \mathrm{C}$, and co-efficient of thermal conductivity of thermacole is $0.01 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$. [Heat of fusion of water $=335 \times 103 \mathrm{~J} \mathrm{~kg}^{-1}$ ].

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Answer

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, $T = 45°C$ Coefficient of thermal conductivity of thermacole, $K = 0.01 J s^{–1}m^{–1}K^{–1}$ Heat of fusion of water, $L = 335 \times 10^3J kg^{–1}$ Let m’ be the total amount of ice that melts in 6h. The amount of heat lost by the food: $\theta=\text{KA}(\text{T}-0)\frac{\text{t}}{1}$ Where,
A = Surface area of the box = $6s^2 = 6 \times (0.3)^2 = 0.54m^3$ $\theta=0.01\times0.54\times45\times6\times60\times\frac{60}{0.05}=104976\text{J}$ But $\theta=\text{m}'\text{L}$ $\therefore\text{m}'=\frac{\theta}{\text{L}}$ $=\frac{104976}{(335\times10^3)}=0.313\text{kg}$ Mass of ice left = 4 – 0.313 = 3.687kg Hence, the amount of ice remaining after 6 h is 3.687kg.

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