A hot body obeying Newton’s law of cooling, cooling down from its peak value $80^{0} C$ to an ambient temperature of $30^{0} C$ . It takes 5 minutes in cooling down form $80^{0} C$ to $40^{0} C$ . How much time will it take to cool down from $60^{0} C$ to $30^{0} C$

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9.6 minutes

7.5 minutes

3.75 minutes

8.6 minutes

answer is D.

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Detailed Solution

$According to Newton' s law of cooling$

$(\frac{θ_{1} - θ_{2}}{t}) = k (\frac{θ_{1} + θ_{2}}{2} - θ_{0})$

$first case \frac{40^{0}}{5} = k (60^{0} - 30^{0}) - - - - - (1)$

$second case \frac{30^{0}}{t} = k (45^{0} - 30^{0}) - - - - (2)$

$dividing \frac{8}{\frac{30}{t}} = 2 t = 7.5 m i n u t e s$

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