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

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(D) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference in temperature between the body and its surroundings: $

Vedclass · Accepted Answer

$3.89$

Vedclass · Answer

(D) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference in temperature between the body and its surroundings: $\frac{d	heta}{dt} = -K(	heta_{avg} - 	heta_s)$.In the first case,the body cools from $70^{\circ} C$ to $40^{\circ} C$ in $5 	ext{ min}$.Average temperature $	heta_{avg1} = \frac{70+40}{2} = 55^{\circ} C$.Excess temperature $= 55 - 20 = 35^{\circ} C$.Rate of cooling $\frac{d	heta_1}{dt} = \frac{70-40}{5} = 6^{\circ} C/	ext{min}$.So,$6 = K 	imes 35 \implies K = \frac{6}{35} \dots (1)$.In the second case,the body cools from $60^{\circ} C$ to $40^{\circ} C$ in time $t$.Average temperature $	heta_{avg2} = \frac{60+40}{2} = 50^{\circ} C$.Excess temperature $= 50 - 20 = 30^{\circ} C$.Rate of cooling $\frac{d	heta_2}{dt} = \frac{60-40}{t} = \frac{20}{t}$.So,$\frac{20}{t} = K 	imes 30 \dots (2)$.Dividing equation $(1)$ by $(2)$:$\frac{6}{20/t} = \frac{35}{30} \implies \frac{6t}{20} = \frac{7}{6}$.$t = \frac{7 	imes 20}{6 	imes 6} = \frac{140}{36} \approx 3.89 	ext{ min}$.

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

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