(A) According to Newton's Law of Cooling,$\frac{dT}{dt} = K(T_{avg} - T_{room})$.For the first interval: $\frac{365 - 361}{2} = K \left( \frac{365 + 3

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(A) According to Newton's Law of Cooling,$\frac{dT}{dt} = K(T_{avg} - T_{room})$.For the first interval: $\frac{365 - 361}{2} = K \left( \frac{365 + 361}{2} - 293 \right)$.$2 = K(363 - 293) = K(70)$.$K = \frac{2}{70} = \frac{1}{35} \ \text{min}^{-1}$.For the second interval: $\frac{344 - 342}{t} = K \left( \frac{344 + 342}{2} - 293 \right)$.$\frac{2}{t} = \frac{1}{35} (343 - 293) = \frac{1}{35} (50) = \frac{10}{7}$.$t = \frac{2 \times 7}{10} = 1.4 \ \text{minutes}$.$t = 1.4 \times 60 = 84 \ \text{seconds}$.

Class 11 Physics 10-2.Heat Transfer Newton's Law of Cooling

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The temperature of a liquid drops from $365 \ K$ to $361 \ K$ in $2$ minutes. Find the time during which the temperature of the liquid drops from $344 \ K$ to $342 \ K$. The temperature of the room is $293 \ K$.

A
$84$
B
$72$
C
$66$
D
$60$

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10-2.Heat Transfer

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Newton's Law of Cooling

$A$ vessel contains hot water at $100^{\circ}C$. If it cools to $80^{\circ}C$ in time $T_1$ and from $80^{\circ}C$ to $60^{\circ}C$ in time $T_2$,then:

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$A$ body cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes when kept in air at $30^{\circ}C$. In the next $10$ minutes,its temperature will be:

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$A$ bucket full of hot water is kept in a room. If it cools from $75^{\circ} C$ to $70^{\circ} C$ in $t_1$ minutes,from $70^{\circ} C$ to $65^{\circ} C$ in $t_2$ minutes and $65^{\circ} C$ to $60^{\circ} C$ in $t_3$ minutes,then

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$A$ metallic sphere cools from $50^{\circ}C$ to $40^{\circ}C$ in $300 \, s$. If the atmospheric temperature is $20^{\circ}C$,then the sphere's temperature after the next $5$ minutes will be close to $.....^{\circ}C$.

MediumJEE MAIN 2020

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$A$ body cools from a temperature $3T$ to $2T$ in $10$ minutes. The room temperature is $T$. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of the next $10$ minutes will be

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The temperature of a liquid drops from $365 \ K$ to $361 \ K$ in $2$ minutes. Find the time during which the temperature of the liquid drops from $344 \ K$ to $342 \ K$. The temperature of the room is $293 \ K$.

Similar Questions

$A$ vessel contains hot water at $100^{\circ}C$. If it cools to $80^{\circ}C$ in time $T_1$ and from $80^{\circ}C$ to $60^{\circ}C$ in time $T_2$,then:

$A$ body cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes when kept in air at $30^{\circ}C$. In the next $10$ minutes,its temperature will be:

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