A 5cm thick ice block is there on the surface of water in a lake. The temperature of air is $- 10^{\circ}$ C; how much time it will take to double the thickness of the block
$(L = 80 c a l / g, K_{i c e} = 0.004 E r g / s - k, d_{i c e} = 0.92 g c m^{- 3})$

1 hour

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191 hours

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19.1 hours

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1.91 hours

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Solution

The correct option is C 19.1 hours
$t = \frac{Q l}{K A (θ_{1} - θ_{2})} = \frac{m L l}{K A (θ_{1} - θ_{2})} = \frac{V ρ L l}{K A (θ_{1} - θ_{2})}$
$= \frac{5 \times A \times 0.92 \times 80 \times \frac{5 + 10}{2}}{0.004 \times A \times 10 \times 3600} = 19. 1 h o u r s .$

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A 5cm thick ice block is there on the surface of water in a lake. The temperature of air is $- 10^{\circ}$ C; how much time it will take to double the thickness of the block
$(L = 80 c a l / g, K_{i c e} = 0.004 E r g / s - k, d_{i c e} = 0.92 g c m^{- 3})$

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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 (L=80 cal/g,Kice=0.004 Erg/s−k,dice=0.92gcm−3)

The correct option is C 19.1 hourst=QlKA(θ1−θ2)=mLlKA(θ1−θ2)=VρLlKA(θ1−θ2) =5×A×0.92×80×5+1020.004×A×10×3600=19. 1 hours.

A 5cm thick ice block is there on the surface of water in a lake. The temperature of air is $- 10^{\circ}$ C; how much time it will take to double the thickness of the block
$(L = 80 c a l / g, K_{i c e} = 0.004 E r g / s - k, d_{i c e} = 0.92 g c m^{- 3})$

The correct option is C 19.1 hours
$t = \frac{Q l}{K A (θ_{1} - θ_{2})} = \frac{m L l}{K A (θ_{1} - θ_{2})} = \frac{V ρ L l}{K A (θ_{1} - θ_{2})}$
$= \frac{5 \times A \times 0.92 \times 80 \times \frac{5 + 10}{2}}{0.004 \times A \times 10 \times 3600} = 19. 1 h o u r s .$