A body cools from 70^{circ} C to 40^{circ} C | vedclass

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(B) According to Newton's law of cooling, the rate of cooling is directly proportional to the temperature difference between the body and the surround

Vedclass · Accepted Answer

$7$

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(B) According to Newton's law of cooling, the rate of cooling is directly proportional to the temperature difference between the body and the surroundings.$\frac{T_i - T_f}{t} = K \left( \frac{T_i + T_f}{2} - T_0 ight)$Where $T_i$ is initial temperature, $T_f$ is final temperature, $t$ is time, $T_0$ is surrounding temperature, and $K$ is a constant.For the first case: $T_i = 70^{\circ} C$, $T_f = 40^{\circ} C$, $t = 5 \ min$, $T_0 = 20^{\circ} C$.$\frac{70 - 40}{5} = K \left( \frac{70 + 40}{2} - 20 ight)$$\frac{30}{5} = K (55 - 20)$$6 = K(35) \Rightarrow K = \frac{6}{35}$For the second case: $T_i = 60^{\circ} C$, $T_f = 30^{\circ} C$, $t = ?$, $T_0 = 20^{\circ} C$.$\frac{60 - 30}{t} = \frac{6}{35} \left( \frac{60 + 30}{2} - 20 ight)$$\frac{30}{t} = \frac{6}{35} (45 - 20)$$\frac{30}{t} = \frac{6}{35} (25)$$t = \frac{30 	imes 35}{6 	imes 25} = \frac{5 	imes 35}{25} = \frac{35}{5} = 7 \ min.$

$A$ body cools from $70^{\circ} C$ to $40^{\circ} C$ in $5$ minutes. Calculate the time it takes to cool from $60^{\circ} C$ to $30^{\circ} C$. The temperature of the surroundings is $20^{\circ} C$. (in $min.$)

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