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A copper pipe of length $10 \,m$ carries steam at temperature $110^{\circ} C$. The outer surface of the pipe is maintained at a temperature $10^{\circ} C$. The inner and outer radii of the pipe are $2 \,cm$ and $4 \,cm$, respectively. The thermal conductivity of copper is $0.38 kW / m /{ }^{\circ} C$. In the steady state, the rate at which heat flows radially outward through the pipe is closest to ............. $\,kW$
$3245$
$3445$
$3645$
$3845$
Solution

(B)
Heat current will flow radially outward.
We know $R _{ T }=\frac{\ell}{ kA }$
Here $\int dR _{ T }=\int \limits_{ R _1}^{ R _2} \frac{ dx }{ k 2 \pi x \ell}$
$\Rightarrow R _{ T }=\frac{1}{2 \pi k \ell} \ln \left(\frac{ R _2}{ R _1}\right)$
$\frac{ dQ }{ dt }=\frac{\Delta T }{ R _{ T }}$
$\therefore \frac{ dQ }{ dt }=\frac{(110-10) 2 \pi(0.38)(10)}{\ell n(2)}$
$=3444.6 \approx 3445 \,kW$