Q. A $5cm$ thick ice block is there on the surface of water in a lake. The temperature of air is $- 10℃$; how much time it will take to double the thickness of the block $(L = 80\ cal/g,\ K_{ice} = 0.004\ erg/s$-$k,d_{ice} = 0.92\ g\ cm^{- 3})$
A) $1\ hour$
B) $191\ hours$
C) $19.1\ hours$
D) $1.91\ hours$
Explanation
(c) $t = \frac{Ql}{KA(\theta_{1} - \theta_{2})} = \frac{mLl}{KA(\theta_{1} - \theta_{2})} = \frac{V\rho Ll}{KA(\theta_{1} - \theta_{2})}$
$= \frac{5 \times A \times 0.92 \times \frac{5 + 10}{2}}{0.004 \times A \times 10 \times 3600} = 19.1hours$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.