A body cools down from $45^{o} C$ to $40^{o} C$ in $5 m i n u t e s$ and to $35^{o}$ in $8 m i n u t e s$ . Find the temperature of the surrounding.

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Solution

According to Newton’s law of cooling:

$\frac{- d T}{d t} = K (T - T_{0})$

Here, temperature changes from $45^{0} C t o 40^{0} C i n 5 m i n u t e s$

So, equation becomes $- \frac{5}{300} = K (45 - T_{0})$

$\frac{- 1}{60} = K (45 - T_{0}) . . . . . . . . . . . . . . (1)$

Now, temperature changes from $40^{0} C t o 35^{0} C i n a n o t h e r 8 min u t e s . s o, 8 minutes - 5 min u t e s = 3 min u t e s$

So, equation becomes

$\frac{- 5}{180} = K (40 - T_{0})$

$- \frac{1}{36} = K (40 - T_{0}) . . . . . . . . . . . . (2)$

Dividing equation (1) and (2)

$\frac{36}{60} = \frac{45 - T_{0}}{40 - T_{0}}$

$\frac{3}{5} = \frac{45 - T_{0}}{40 - T_{0}}$

$2 T_{0} = 105$

$T_{0} = {52.5}^{0} C$

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