Question

A metal rod of length $10 \,cm$ and area of cross-section $2.8 \times 10^{-4} m ^{2}$ is covered with a non-conducting substance. One end of it is maintained at $80^{\circ} C$, while the other end is put in ice at $0^{\circ} C .$ It is found that $20\, gm$ of ice melts in $5\, min$. The thermal conductivity of the metal in $Js ^{-1} m ^{-1} K ^{-1}$ is (Latent heat of ice is $80 \,cal\, g ^{-1} .$ )

• A. 70
• B. 80
• C. 90
• D. 100

Answer 1

Answer: (D)

Given,
length of rod, $l=10 \,cm =0.1 \,m$
area of cross-section of rod, $A=2.8 \times 10^{-4}\, m ^{2}$
temperatuer at one end, $T_{1}=80^{\circ} C$
temperature at other end, $T_{2}=0^{\circ} C$
quantity of melted ice, $m =20 \,gm$
time taken to melt ice, $=5 \,min =300 \,sec$
and latent heat of ice, $s=80\, cal \,g ^{-1}$
Now, rate of the heat flow $=\frac{m \times s \times 4.184}{t}$
$=\frac{20 \times 80 \times 4.184}{300}=22.314 J / s$
Rate of heat flow in the rod, $\Delta Q=\frac{k A \cdot \Delta T}{l}$
$22.314=\frac{k\left(2.8 \times 10^{4}\right) \times 80}{0.1} $
$\therefore k=99.61 \approx 100\, Js ^{-1} \,m ^{-1} \,K ^{-1}$