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(A) Side of the cubical ice box,$s = 30 \,cm = 0.3 \,m$.Thickness of the ice box,$l = 5.0 \,cm = 0.05 \,m$.Mass of ice initially,$m_{initial} = 4 \,kg

Vedclass · Accepted Answer

$3.69$

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(A) Side of the cubical ice box,$s = 30 \,cm = 0.3 \,m$.Thickness of the ice box,$l = 5.0 \,cm = 0.05 \,m$.Mass of ice initially,$m_{initial} = 4 \,kg$.Time duration,$t = 6 \,h = 6 	imes 3600 \,s = 21600 \,s$.Temperature difference,$\Delta T = 45 \,^{\circ}C - 0 \,^{\circ}C = 45 \,K$.Coefficient of thermal conductivity,$K = 0.01 \,J \,s^{-1} \,m^{-1} \,K^{-1}$.Latent heat of fusion of water,$L = 335 	imes 10^{3} \,J \,kg^{-1}$.Surface area of the cubical box,$A = 6s^{2} = 6 	imes (0.3)^{2} = 0.54 \,m^{2}$.The heat conducted into the box is given by $Q = \frac{K A \Delta T t}{l}$.$Q = \frac{0.01 	imes 0.54 	imes 45 	imes 21600}{0.05} = 104976 \,J$.The mass of ice that melts,$m_{melt} = \frac{Q}{L} = \frac{104976}{335 	imes 10^{3}} \approx 0.313 \,kg$.The remaining mass of ice is $m_{remaining} = m_{initial} - m_{melt} = 4 - 0.313 = 3.687 \,kg \approx 3.69 \,kg$.

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