(D) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{surr})$.For the first case: $\frac{65.5 - 62.

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(D) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{surr})$.For the first case: $\frac{65.5 - 62.5}{1} = K \left( \frac{65.5 + 62.5}{2} - 22.5 \right)$.$3 = K(64 - 22.5) = K(41.5)$,so $K = \frac{3}{41.5}$.For the second case: $\frac{46.5 - 40.5}{t} = K \left( \frac{46.5 + 40.5}{2} - 22.5 \right)$.$\frac{6}{t} = \frac{3}{41.5} (43.5 - 22.5) = \frac{3}{41.5} (21)$.$t = \frac{6 \times 41.5}{3 \times 21} = \frac{2 \times 41.5}{21} = \frac{83}{21} \approx 3.95 \approx 4$ minutes.

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

Difficult

$A$ cup of tea cools from $65.5 ^\circ C$ to $62.5 ^\circ C$ in $1$ minute in a room at $22.5 ^\circ C$. How long will the same cup of tea take,in minutes,to cool from $46.5 ^\circ C$ to $40.5 ^\circ C$ in the same room? (Choose the nearest value)

A
$1$
B
$2$
C
$3$
D
$4$

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10-2.Heat Transfer

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Newton's Law of Cooling

The instantaneous temperature difference between a cooling body and its surroundings,which obeys Newton's law of cooling,is $\theta$. Which of the following graphs represents the variation of $\ln \theta$ with time $t$?

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$A$ hollow copper sphere $S$ and a hollow copper cube $C$,both with negligible thin walls of the same surface area,are filled with water at $90^{\circ}C$ and allowed to cool in the same environment. The graph that correctly represents their cooling is:

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For a system where Newton's law of cooling is applicable,the initial rate of cooling is $R \ ^\circ C/sec$. Find the time when the temperature difference $\Delta T_0$ (initial temperature difference) is reduced to half.

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Describe the procedure to demonstrate that the rate of loss of heat from a hot body is directly dependent on the temperature difference between the body and its surroundings.

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Hot water cools from $60\,^{\circ}C$ to $50\,^{\circ}C$ in the first $10\,minutes$ and to $42\,^{\circ}C$ in the next $10\,minutes$. The temperature of the surroundings is ......... $^{\circ}C$

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$A$ cup of tea cools from $65.5 ^\circ C$ to $62.5 ^\circ C$ in $1$ minute in a room at $22.5 ^\circ C$. How long will the same cup of tea take,in minutes,to cool from $46.5 ^\circ C$ to $40.5 ^\circ C$ in the same room? (Choose the nearest value)

Similar Questions

The instantaneous temperature difference between a cooling body and its surroundings,which obeys Newton's law of cooling,is $\theta$. Which of the following graphs represents the variation of $\ln \theta$ with time $t$?

$A$ hollow copper sphere $S$ and a hollow copper cube $C$,both with negligible thin walls of the same surface area,are filled with water at $90^{\circ}C$ and allowed to cool in the same environment. The graph that correctly represents their cooling is:

For a system where Newton's law of cooling is applicable,the initial rate of cooling is $R \ ^\circ C/sec$. Find the time when the temperature difference $\Delta T_0$ (initial temperature difference) is reduced to half.

Describe the procedure to demonstrate that the rate of loss of heat from a hot body is directly dependent on the temperature difference between the body and its surroundings.

Hot water cools from $60\,^{\circ}C$ to $50\,^{\circ}C$ in the first $10\,minutes$ and to $42\,^{\circ}C$ in the next $10\,minutes$. The temperature of the surroundings is ......... $^{\circ}C$

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