A body cools from 80,^{circ} C to 50,^{circ} | vedclass

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(D) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings:$-\fr

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$10$

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(D) According to Newton's law of cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings:$-\frac{dT}{dt} = K(T - T_0)$Integrating this,we get $\ln\left(\frac{T_1 - T_0}{T_2 - T_0}\right) = K \cdot t$.For the first case:$T_1 = 80\,^{\circ} C$,$T_2 = 50\,^{\circ} C$,$T_0 = 20\,^{\circ} C$,$t = 5\, min$.$\ln\left(\frac{80 - 20}{50 - 20}\right) = K \cdot 5$$\ln\left(\frac{60}{30}\right) = 5K \implies \ln(2) = 5K \implies K = \frac{\ln(2)}{5}$.For the second case:$T_1 = 60\,^{\circ} C$,$T_2 = 30\,^{\circ} C$,$T_0 = 20\,^{\circ} C$,$t = ?$$\ln\left(\frac{60 - 20}{30 - 20}\right) = K \cdot t$$\ln\left(\frac{40}{10}\right) = K \cdot t$$\ln(4) = K \cdot t$Since $\ln(4) = 2 \ln(2)$:$2 \ln(2) = \left(\frac{\ln(2)}{5}\right) \cdot t$$2 = \frac{t}{5} \implies t = 10\, min$.

$A$ body cools from $80\,^{\circ} C$ to $50\,^{\circ} C$ in $5$ minutes. Calculate the time (in $min$) it takes to cool from $60\,^{\circ} C$ to $30\,^{\circ} C$. The temperature of the surroundings is $20\,^{\circ} C$.

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