If K_{1} and K_{2} are the thermal conductivi | vedclass

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(C) In the steady state,the rate of heat flow through the steel rod must be equal to the rate of heat flow through the copper rod.$\frac{dQ}{dt} = \fr

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) In the steady state,the rate of heat flow through the steel rod must be equal to the rate of heat flow through the copper rod.$\frac{dQ}{dt} = \frac{K_{1} A_{1} (T_{1} - T)}{L_{1}} = \frac{K_{2} A_{2} (T - T_{2})}{L_{2}}$Given: $T_{1} = 450^{\circ}C$,$T_{2} = 0^{\circ}C$,$\frac{K_{2}}{K_{1}} = 9$,$\frac{A_{1}}{A_{2}} = 2$,$\frac{L_{1}}{L_{2}} = 2$.Substituting these values into the equation:$\frac{450 - T}{T - 0} = \frac{K_{2}}{K_{1}} 	imes \frac{A_{2}}{A_{1}} 	imes \frac{L_{1}}{L_{2}}$$\frac{450 - T}{T} = 9 	imes \frac{1}{2} 	imes 2 = 9$$450 - T = 9T$$10T = 450$$T = 45^{\circ}C$

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