Question

A metal cylinder $0.628m$ long and $0.04m$ in diameter has one end in boiling water at ${100}^{o}C$ and other end is melting ice. The co-efficient of Thermal conductivity of the metal is $378W{m}^{-1}{k}^{-1}$. Latest heat of Ice is $3.36\times {10}^{5}Jk{g}^{-1}$. Find the mass of ice melts in one hour.

Answer 1

$l=0.628meter$

$A=\pi r^2=\pi \frac{d^2}{r}=\frac{\pi }{4}\times (0.04{)}^2=4\pi \times {10}^{-4}{m}^2$

$\Delta \theta =100-0={100}^{o}C=373k-273k=100k$

$k=378\frac{W}{mKelvin}$

$L=3.36\times {10}^{5}J/kg$

one hour $t=60\times 60second=3600sec$

so heat fastered in $t$ second will be

$Q=kA\frac{\Delta Q}{l}.t=378\times \frac{4\pi \times {10}^{-4}\times 100}{0.628}\times 3600$

or $Q=2.7216\times {10}^{9}Joule$

ice melt $m=\frac{Q}{L}=\frac{2.7216\times {10}^9}{3.36\times {10}^5}=8.1\times {10}^{3}kg$