Question

It takes 3 hours to freeze the lake from 2.1 cm to 2.4cm. How long will it take to freeze it from 3.6cm to 3.9cm.

Solution

Time taken to freeze the lake from thickness x₁to x₂ is given by,

$straight t space equals space fraction numerator ρL over denominator 2 Kθ end fraction space left parenthesis straight x subscript 2 squared space minus space straight x subscript 1 squared right parenthesis thin space straight t space proportional to space space left parenthesis straight x subscript 2 squared space minus space straight x subscript 1 squared right parenthesis thin space fraction numerator straight t apostrophe over denominator straight t end fraction space equals space fraction numerator left parenthesis straight x apostrophe subscript 2 squared space minus space straight x apostrophe subscript 1 squared right parenthesis over denominator straight x subscript 2 squared space minus space straight x subscript 1 squared end fraction fraction numerator straight t apostrophe over denominator 3 end fraction space equals space fraction numerator left parenthesis 3.9 right parenthesis squared space minus space left parenthesis 3.6 right parenthesis squared over denominator left parenthesis 2.4 right parenthesis squared space minus space left parenthesis 2.1 right parenthesis squared end fraction space equals space 5 over 3 straight t apostrophe space equals space 5 space hours$

Some More Questions From Mechanical Properties of Fluids Chapter

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Mechanical Properties Of Fluids

It takes 3 hours to freeze the lake from 2.1 cm to 2.4cm. How long will it take to freeze it from 3.6cm to 3.9cm.

Some More Questions From Mechanical Properties of Fluids Chapter

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