Question

Derive an expression for the time taken to cool a body from temperature T₁ to T₂placed in an enclosure at temperature T₀

Solution

According to Newton’s law of cooling,

space space space space fraction numerator space dT over denominator dt end fraction equals negative open parentheses 4 eAσ over ms straight T subscript 0 superscript 3 close parentheses left parenthesis straight T minus straight T subscript 0 right parenthesis rightwards double arrow space space space space space dt equals negative open parentheses 4 eAσ over ms straight T subscript 0 superscript 3 close parentheses fraction numerator dT over denominator straight T minus straight T subscript 0 end fraction

Let it take time 't' for the temperature to fall from T₁ to T_2.On integrating the above equation,

integral subscript 0 superscript straight t dT space equals space minus open parentheses 4 eAσ over ms straight T subscript straight o cubed close parentheses space integral subscript straight T subscript 1 end subscript superscript straight T subscript 2 end superscript space fraction numerator dT over denominator straight T space minus space straight T subscript straight o end fraction space rightwards double arrow space straight t space equals space open parentheses 4 eAσ over ms straight T subscript straight o cubed close parentheses space Ln space open parentheses fraction numerator straight T subscript 1 space minus space straight T subscript straight o over denominator straight T subscript 2 space minus space straight T subscript straight o end fraction close parentheses rightwards double arrow space straight t space proportional to space Ln space open parentheses fraction numerator straight T subscript 1 space minus space straight T subscript straight o over denominator straight T subscript 2 space minus space straight T subscript straight o end fraction close parentheses

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