A cylindrical copper rod of length 2 ,m and c | vedclass

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(C) Given:Length of the rod $L = 2 \,m = 200 \,cm$.Temperature at the hot end $	heta_1 = 100^{\circ} C$.Temperature at the cold end $	heta_2 = 0^{\c

Vedclass · Accepted Answer

$60$

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(C) Given:Length of the rod $L = 2 \,m = 200 \,cm$.Temperature at the hot end $	heta_1 = 100^{\circ} C$.Temperature at the cold end $	heta_2 = 0^{\circ} C$.Distance from the cold end $x_2 = 120 \,cm$.Distance from the hot end $x_1 = L - x_2 = 200 \,cm - 120 \,cm = 80 \,cm$.At steady state,the rate of heat flow $(dQ/dt)$ through any cross-section of the rod is constant.Since $dQ/dt = KA(\Delta 	heta / \Delta x)$,and $K$ and $A$ are constant for the rod,the temperature gradient $(\Delta 	heta / \Delta x)$ must be constant throughout the rod.Therefore,$\frac{	heta_1 - 	heta}{x_1} = \frac{	heta - 	heta_2}{x_2}$.Substituting the values:$\frac{100^{\circ} C - 	heta}{80 \,cm} = \frac{	heta - 0^{\circ} C}{120 \,cm}$.$120(100 - 	heta) = 80	heta$.$12000 - 120	heta = 80	heta$.$200	heta = 12000$.$	heta = \frac{12000}{200} = 60^{\circ} C$.Thus,the temperature at the point is $60^{\circ} C$.

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