A $5cm$ thick ice block is there on the surface of water in a lake. The temperature of air is $-10°C$ ; how much time it will take to double the thickness of the block ...... hour ($L = 80 cal/g, Kicc = 0.004 Erg/s-k, dice = 0.92 g cm^{-3}$)
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(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 80 \times \frac{{5 + 10}}{2}}}{{0.004 \times A \times 10 \times 3600}} = 19.1\,hours$.
$ = \frac{{5 \times A \times 0.92 \times 80 \times \frac{{5 + 10}}{2}}}{{0.004 \times A \times 10 \times 3600}} = 19.1\,hours$.
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