(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_s)$,where $T_s$ is the surrounding temperature

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(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_s)$,where $T_s$ is the surrounding temperature.For the first interval: $\frac{50.0 - 49.9}{5} = K \left( \frac{50.0 + 49.9}{2} - 30.0 \right) \implies \frac{0.1}{5} = K(49.95 - 30.0) = K(19.95)$ ... $(i)$For the second interval: $\frac{40.0 - 39.9}{t} = K \left( \frac{40.0 + 39.9}{2} - 30.0 \right) \implies \frac{0.1}{t} = K(39.95 - 30.0) = K(9.95)$ ... $(ii)$Dividing equation $(i)$ by $(ii)$:$\frac{0.1/5}{0.1/t} = \frac{K(19.95)}{K(9.95)} \implies \frac{t}{5} = \frac{19.95}{9.95} \approx 2$$t = 5 \times 2 = 10\;s$.

Class 11 Physics 2 Thermal Properties of Matter: Heat Transfer Newton's Law of Cooling

Medium

$A$ body cools from $50.0^{\circ}C$ to $49.9^{\circ}C$ in $5\;s$. How long will it take to cool from $40.0^{\circ}C$ to $39.9^{\circ}C$? Assume the temperature of surroundings to be $30.0^{\circ}C$ and Newton's law of cooling to be valid. The time taken is ....... $s$.

A
$2.5$
B
$10$
C
$20$
D
$5$

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2 Thermal Properties of Matter: Heat Transfer

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Newton's Law of Cooling

Two solid spheres $A$ and $B$ made of the same material have radii $r_A$ and $r_B$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature of $A$ and $B$ is :

DifficultTS EAMCET 2006

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$A$ hot body is allowed to cool. The surrounding temperature is constant at $30^{\circ} C$. It takes time $t_{1}$ to cool from $70^{\circ} C$ to $68^{\circ} C$ and time $t_{2}$ to cool from $60^{\circ} C$ to $59.5^{\circ} C$. Then:

MediumKCET 2012

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$A$ sphere having temperature $600 \text{ K}$ is losing heat due to radiation. At this temperature its rate of cooling is $R$. The rate of cooling of this sphere at $400 \text{ K}$ is $\frac{x}{243}R$. The value of $x$ is (Temperature of surrounding is $300 \text{ K}$)

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$A$ container is filled with a liquid that cools from $100^{\circ}C$ to $70^{\circ}C$ in $5 \text{ min}$, when kept at a room temperature of $30^{\circ}C$. The time that it must have taken to cool down to $80^{\circ}C$ from its initial temperature is approximately: (in $\text{ min}$)

EasyTS EAMCET 2018

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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 $293 \ K$.

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$A$ body cools from $50.0^{\circ}C$ to $49.9^{\circ}C$ in $5\;s$. How long will it take to cool from $40.0^{\circ}C$ to $39.9^{\circ}C$? Assume the temperature of surroundings to be $30.0^{\circ}C$ and Newton's law of cooling to be valid. The time taken is ....... $s$.

Similar Questions

Two solid spheres $A$ and $B$ made of the same material have radii $r_A$ and $r_B$ respectively. Both the spheres are cooled from the same temperature under the conditions valid for Newton's law of cooling. The ratio of the rate of change of temperature of $A$ and $B$ is :

$A$ hot body is allowed to cool. The surrounding temperature is constant at $30^{\circ} C$. It takes time $t_{1}$ to cool from $70^{\circ} C$ to $68^{\circ} C$ and time $t_{2}$ to cool from $60^{\circ} C$ to $59.5^{\circ} C$. Then:

$A$ sphere having temperature $600 \text{ K}$ is losing heat due to radiation. At this temperature its rate of cooling is $R$. The rate of cooling of this sphere at $400 \text{ K}$ is $\frac{x}{243}R$. The value of $x$ is (Temperature of surrounding is $300 \text{ K}$)

$A$ container is filled with a liquid that cools from $100^{\circ}C$ to $70^{\circ}C$ in $5 \text{ min}$, when kept at a room temperature of $30^{\circ}C$. The time that it must have taken to cool down to $80^{\circ}C$ from its initial temperature is approximately: (in $\text{ min}$)

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 $293 \ K$.

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