Question

In the steady state of temperature at the end of a copper rod of length $25cm$ and area of cross-section $1.0{cm}^2$ are ${125}^{o}C$ and $0^{o}C$ respectively. Calculate the temperature gradient, rate of heat flow and the temperature at a distance $10cm$ from the hot end. [$K$ for copper $=9.2\times {10}^{-3}kilocal/m^{o}C-s$]

Answer 1

$l=25cm=0.25mA=1c{m}^2=1\times {10}^{-4}{m}^{2}K=9.2\times {10}^{-3}kcal/m^{o}Cs$

$\frac{125-0}{\frac{25\times {10}^{-2}}{9.2\times {10}^{-3}\times 1\times {10}^{-4}}}=\frac{x-0}{\frac{15\times {10}^{-2}}{9.2\times {10}^{-3}\times 1\times {10}^{-4}}}\Rightarrow \frac{125}{25}=\frac{x}{15}\Rightarrow x=75℃$

Temperature gradient (at 10cm from hotter end)

$=\frac{75-0}{0.15}K/m=500K/m$