A certain quantity of water cools from $70^{\circ} ext{C}$ to $60^{\circ} ext{C}$ in the first $5 ext{ minu

Question

A certain quantity of water cools from $70^{\circ}	ext{C}$ to $60^{\circ}	ext{C}$ in the first $5	ext{ minutes}$ and to $54^{\circ}	ext{C}$ in the next $5	ext{ minutes}$. The temperature of the surroundings will be:

Accepted Answer

According to the average form of Newton's law of cooling:
$\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s ight)$
where $T_s$ is the temperature of the surroundings.

For the first $5	ext{ minutes}$:
$\frac{70 - 60}{5} = K \left( \frac{70 + 60}{2} - T_s ight)$
$2 = K(65 - T_s)$  ... (i)

For the next $5	ext{ minutes}$:
$\frac{60 - 54}{5} = K \left( \frac{60 + 54}{2} - T_s ight)$
$1.2 = K(57 - T_s)$ ... (ii)

Dividing equation (i) by (ii):
$\frac{2}{1.2} = \frac{65 - T_s}{57 - T_s}$
$\frac{5}{3} = \frac{65 - T_s}{57 - T_s}$
$5(57 - T_s) = 3(65 - T_s)$
$285 - 5T_s = 195 - 3T_s$
$2T_s = 90$
$T_s = 45^{\circ}	ext{C}$

Therefore, the temperature of the surroundings is $45^{\circ}	ext{C}$.

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