Question

A metal rod of length $10cm$ 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 $20gm$ of ice melts in $5min$. The thermal conductivity of the metal in $J{s}^{-1}{m}^{-1}{K}^{-1}$ is (Latent heat of ice is $80cal{g}^{-1}.$ )

• A. 70
• B. 80
• C. 90
• D. 100

Answer 1

Answer: (D)

Given,
length of rod, $l=10cm=0.1m$
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, $m=20gm$
time taken to melt ice, $=5min=300sec$
and latent heat of ice, $s=80cal{g}^{-1}$
Now, rate of the heat flow $=\frac{m\times s\times 4.184}{t}$
$=\frac{20\times 80\times 4.184}{300}=22.314J/s$
Rate of heat flow in the rod, $\Delta Q=\frac{kA\cdot \Delta T}{l}$
$22.314=\frac{k\left(2.8\times 10^4\right)\times 80}{0.1}$
$\therefore k=99.61\approx 100J{s}^{-1}{m}^{-1}{K}^{-1}$