The walls of a closed cubical box of edge 60 | vedclass

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(C) The rate of heat flow through the walls of the cube by conduction is given by: $\frac{dQ}{dt} = \frac{kA(T_1 - T_0)}{x}$.Here,the total surface ar

Vedclass · Accepted Answer

$0.441$

Vedclass · Answer

(C) The rate of heat flow through the walls of the cube by conduction is given by: $\frac{dQ}{dt} = \frac{kA(T_1 - T_0)}{x}$.Here,the total surface area $A = 6a^2 = 6 	imes (60 	ext{ cm})^2 = 6 	imes 3600 = 21600 	ext{ cm}^2$.The thickness $x = 1 	ext{ mm} = 0.1 	ext{ cm}$.The thermal conductivity $k = 4 	imes 10^{-4} 	ext{ cal s}^{-1} 	ext{ cm}^{-1} {}^{\circ}	ext{C}^{-1}$.The temperature difference $\Delta T = 1000^{\circ}	ext{C}$.Converting the heat flow rate to $SI$ units (Watts),we multiply by $4.184 	ext{ J/cal}$:$P = \frac{k A \Delta T}{x} 	imes 4.184 = \frac{4 	imes 10^{-4} 	imes 21600 	imes 1000}{0.1} 	imes 4.184 	ext{ W}$.$P = (4 	imes 21600) 	imes 4.184 = 86400 	imes 4.184 \approx 361497.6 	ext{ W}$ (Wait,recalculating: $4 	imes 10^{-4} 	imes 21600 	imes 1000 / 0.1 = 86400 	ext{ cal/s}$. $86400 	imes 4.184 = 361497.6 	ext{ W}$).Given $P = V^2/R$,then $R = V^2/P = (400)^2 / 361497.6 = 160000 / 361497.6 \approx 0.4426 \Omega$.Rounding to the nearest option,the resistance is $0.441 \Omega$.

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