Three rods of Copper,Brass,and Steel are weld | vedclass

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(B) The rate of heat flow $Q$ is given by the formula:$Q = \frac{KA(	heta_1 - 	heta_2)}{l}$where $K$ is the coefficient of thermal conductivity,$l$

Vedclass · Accepted Answer

$4.8$

Vedclass · Answer

(B) The rate of heat flow $Q$ is given by the formula:$Q = \frac{KA(	heta_1 - 	heta_2)}{l}$where $K$ is the coefficient of thermal conductivity,$l$ is the length of the rod,and $A$ is the area of cross-section.Let $T$ be the temperature at the junction where the three rods meet.According to the principle of conservation of energy,the heat flowing into the junction through the copper rod must equal the heat flowing out through the brass and steel rods:$Q_{	ext{copper}} = Q_{	ext{brass}} + Q_{	ext{steel}}$Substituting the given values:$\frac{0.92 	imes 4 	imes (100 - T)}{46} = \frac{0.26 	imes 4 	imes (T - 0)}{13} + \frac{0.12 	imes 4 	imes (T - 0)}{12}$Simplifying the equation:$0.08 	imes (100 - T) = 0.08 	imes T + 0.04 	imes T$$8 - 0.08T = 0.12T$$8 = 0.2T$$T = \frac{8}{0.2} = 40^\circ C$Now,calculate the rate of heat flow through the copper rod:$Q_{	ext{copper}} = \frac{0.92 	imes 4 	imes (100 - 40)}{46} = \frac{0.92 	imes 4 	imes 60}{46} = 0.02 	imes 4 	imes 60 = 4.8 \ cal \ s^{-1}$.

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