A metal rod of length \(10 \mathrm{~cm}\) and area of cross-section \(2.8 \times 10^{-4} \mathrm{~m}^2\) is…
A metal rod of length \(10 \mathrm{~cm}\) and area of cross-section \(2.8 \times 10^{-4} \mathrm{~m}^2\) is covered with a non-conducting substance. One end of it is maintained at \(80^{\circ} \mathrm{C}\), while the other end is put in ice at \(0^{\circ} \mathrm{C}\). It is found that \(20 \mathrm{gm}\) of ice melts in \(5 \mathrm{~min}\). The thermal conductivity of the metal in Js" \(\mathrm{m}^{-1} \mathrm{~K}^{-1}\) is (Latent heat of ice is \(80 \mathrm{cal} \mathrm{g}^{-1}\).)
70
80
90
100
Solution
Given,
length of rod, \(l=10 \mathrm{~cm}=0.1 \mathrm{~m}\)
area of cross-section of rod, \(A=2.8 \times 10^{-4} \mathrm{~m}^2\)
temperature at one end, \(T_1=80^{\circ} \mathrm{C}\)
temperature at other end, \(T_2=0^{\circ} \mathrm{C}\)
quantity of melted ice, \(\mathrm{m}=20 \mathrm{gm}\)
time taken to melt ice, \(t=5 \mathrm{~min}=300 \mathrm{sec}\)
and latent heat of ice, \(s=80 \mathrm{cal} \mathrm{g}^{-1}\)
Now, rate of the heat flow \(=\frac{m \times s \times 4.184}{t}\)
\(=\frac{20 \times 80 \times 4.184}{300}=22.314 \mathrm{~J} / \mathrm{s}\)
Rate of heat flow in the rod, \(\Delta Q=\frac{k A \cdot \Delta T}{l}\)
\(\begin{aligned}
& 22.314=\frac{k\left(2.8 \times 10^{-4}\right) \times 80}{0.1} \\
\therefore \quad & k=99.61 \approx 100 \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}
\end{aligned}\)