Three metal rods made of copper, brass, and s | vedclass

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(C) Let the temperature of the junction be $T$. The rate of heat flow is given by $H = \frac{KA(T_1 - T_2)}{L}$.Since the rods are joined at a junctio

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$4.8$

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(C) Let the temperature of the junction be $T$. The rate of heat flow is given by $H = \frac{KA(T_1 - T_2)}{L}$.Since the rods are joined at a junction, the sum of heat currents flowing away from the junction must be zero: $H_{Cu} + H_{Br} + H_{St} = 0$.Given: $A = 4 \,cm^2$ for all rods.$H_{Cu} = \frac{0.92 	imes 4 	imes (T - 100)}{46} = 0.08(T - 100)$$H_{Br} = \frac{0.26 	imes 4 	imes (T - 0)}{13} = 0.08T$$H_{St} = \frac{0.12 	imes 4 	imes (T - 0)}{12} = 0.04T$Summing these: $0.08(T - 100) + 0.08T + 0.04T = 0$$0.08T - 8 + 0.08T + 0.04T = 0$$0.20T = 8 \implies T = 40 \,^{\circ}C$.The rate of heat flow through the copper rod is $H_{Cu} = 0.08(40 - 100) = 0.08(-60) = -4.8 \,cal \,s^{-1}$.The magnitude of the rate of heat flow is $4.8 \,cal \,s^{-1}$.

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