There is ice formation in a tank of water of thickness 10cm. How much time it will take to have a layer of 0.1 cm below it? The outer temperature is -5∘C, the thermal conductivity of ice K=0.005 cal/cm-sec∘C, the latent heat of ice is 80 cal/gm and the density of ice is 0.91 gm/cc:

t=ρl2kθy22-y12=0.91×802×0.005×5(10.1)2-(10)2=2926sec=48.77 min.

There is ice formation in a tank of water of thickness 10cm. How much time it will take to have a layer of 0.1 cm below it? The outer temperature is -5∘C, the thermal conductivity of ice K=0.005 cal/cm-sec∘C, the latent heat of ice is 80 cal/gm and the density of ice is 0.91 gm/cc:

t=ρl2kθy22-y12=0.91×802×0.005×5(10.1)2-(10)2=2926sec=48.77 min.

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There is ice formation in a tank of water of thickness $10$ cm. How much time it will take to have a layer of $0.1$ cm below it? The outer temperature is $- 5^{\circ} C$ , the thermal conductivity of ice $K = 0.005 c a l / c m - s e c^{\circ} C$ , the latent heat of ice is $80$ cal/gm and the density of ice is $0.91 g m / c c :$

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$46.39 m i n u t e s$

$47.63 m i n u t e s$

$48.77 m i n u t e s$

$49.31 m i n u t e s$

answer is C.

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Detailed Solution

$t = \frac{ρ l}{2 k θ} (y_{2}^{2} - y_{1}^{2})$

$= \frac{0.91 \times 80}{2 \times 0.005 \times 5} [(10.1)^{2} - (10)^{2}] = 2926 \sec = 48.77 \min .$

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There is ice formation in a tank of water of thickness 10cm. How much time it will take to have a layer of 0.1 cm below it? The outer temperature is -5∘C, the thermal conductivity of ice K=0.005 cal/cm-sec∘C, the latent heat of ice is 80 cal/gm and the density of ice is 0.91 gm/cc:

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