A body initially at temperature 80°C takes 10 minutes to cool to temperature 72°C when the room temperature is 30°C. How long will it take to cool from 65°C to 60°C?

Solution

Time taken by the body to cool from temperature T₁ to T₂ when surrounding temperature is T_o is given by,

space space space space space space space space space space straight t space equals space kLn open parentheses fraction numerator straight T subscript 1 minus straight T subscript straight o over denominator straight T subscript 2 minus straight T subscript straight o end fraction close parentheses

Time taken to cool from 80°C to 72°C is 10 min is,
∴

space space 10 equals kLn open parentheses fraction numerator 80 minus 30 over denominator 72 minus 30 end fraction close parentheses space equals straight k left parenthesis 0.17435 right parenthesis

...(1)
Let t be the time taken to cool from 65°C to 60°C.
∴

straight t equals kLn open parentheses fraction numerator 65 minus 30 over denominator 60 minus 30 end fraction close parentheses space equals space straight k left parenthesis 0.15415 right parenthesis

...(2)
From (1) and (2),

straight t over 10 equals fraction numerator 0.15415 over denominator 0.17435 end fraction

rightwards double arrow

straight t equals 8.84 space min

Therefore, time taken to cool from 65^o C to 60^o C is 8.84 minutes.

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