One end of a uniform metal rod of length 100 | vedclass

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Question

(A) Let the thermal conductivity of the rod be $K$ and the cross-sectional area be $A$. The rate of heat flow is given by $H = \frac{KA \Delta T}{L}$.

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) Let the thermal conductivity of the rod be $K$ and the cross-sectional area be $A$. The rate of heat flow is given by $H = \frac{KA \Delta T}{L}$.For the section in ice (length $x_1 = 60 \, cm$): The temperature difference is $\Delta T_1 = 325^{\circ} C - 0^{\circ} C = 325^{\circ} C$. The rate of heat flow is $H_1 = \frac{KA(325)}{60}$.This heat melts the ice: $H_1 = m_{ice} L_f$, where $L_f$ is the latent heat of fusion.For the section in boiling water (length $x_2 = 100 - 60 = 40 \, cm$): The temperature difference is $\Delta T_2 = 325^{\circ} C - 100^{\circ} C = 225^{\circ} C$. However, heat flows from $325^{\circ} C$ to $100^{\circ} C$, so $H_2 = \frac{KA(325 - 100)}{40} = \frac{KA(225)}{40}$.This heat converts water to steam: $H_2 = m_{steam} L_v$, where $L_v = 6.75 L_f$.Given $m_{steam} = 2 \, g/s$, then $H_2 = 2 	imes 6.75 L_f = 13.5 L_f$.Equating $H_2$: $\frac{KA(225)}{40} = 13.5 L_f$ implies $KA = \frac{13.5 	imes 40}{225} L_f = 2.4 L_f$.Now substitute $KA$ into $H_1$: $H_1 = \frac{2.4 L_f 	imes 325}{60} = 0.04 	imes 325 L_f = 13 L_f$.Since $H_1 = m_{ice} L_f$, we get $m_{ice} = 13 \, g/s$.

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