Question

One end of a lagged cylindrical metal rod of length $31.4cm$ and radius $2cm$ is kept in contact with ice at $0^{\circ }c$ and the other end in water at ${100}^{\circ }C$. Find the rate of melting of ice per minute. $L=80cal/g,k={10}^5$ S.I. units $\left(7.5g\right)$.

Answer 1

Given: One end of a lagged cylindrical metal rod of length 31.4 cm and radius 2 cm is kept in contact with ice at $0^{\circ }c$ and the other end in water at ${100}^{\circ }C$.

To find the rate of melting of ice per minute.

Solution:

As per the given condition,

Length of the rod, $l=31.4cm=0.314m$

radius of the rod, $r=2cm=0.02m$

Area of cross section,$A=4\pi r^2=4\times \pi \times (0.02{)}^2=5\times {10}^{–3}{m}^2$

Thermal coefficient, $k=10^5W/(mK)

Temperature difference between two ends of the rod $(T_1–T_2)=(100–0)=100$

time, $t=1min=60s$

If $\frac{dQ}{dt}$ is the rate of melting of ice per minute, then

We know,

$\frac{dQ}{dt}=\frac{KAt(T_1-T_2)}{l}⟹\frac{dQ}{dt}=\frac{{10}^5\times 5\times {10}^{-3}\times 60\times 100)}{0.314}⟹\frac{dQ}{dt}=955\times {10}^{4}J/s$

is the rate of melting of ice per minute.