(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 circuit below is used to heat water kept in a bucket. Assuming heat loss only by Newton's law of cooling, the variation in the temperature of the water in the bucket as a function of time is depicted by

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The temperature of a liquid in a beaker at time $t$ is $\theta(t)$ and the temperature of the surroundings is $\theta_0$. According to Newton's law of cooling,which of the following is the correct graph between $\log_e(\theta - \theta_0)$ and $t$?

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$A$ body cools from $75 \, ^\circ \text{C}$ to $65 \, ^\circ \text{C}$ in $2 \, \text{min}$ in a room at $30 \, ^\circ \text{C}$. How long will it take to cool from $55 \, ^\circ \text{C}$ to $45 \, ^\circ \text{C}$ in the same room?

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$A$ liquid in a beaker has temperature $\theta(t)$ at time $t$ and $\theta_0$ is the temperature of the surroundings. According to Newton's law of cooling,which of the following is the correct graph between $\log_e(\theta - \theta_0)$ and $t$?

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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 $39.9^{\circ}C$ (in $, s$)? The temperature of the surroundings is $30^{\circ}C$. Assume Newton's Law of Cooling applies.

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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 circuit below is used to heat water kept in a bucket. Assuming heat loss only by Newton's law of cooling, the variation in the temperature of the water in the bucket as a function of time is depicted by

The temperature of a liquid in a beaker at time $t$ is $\theta(t)$ and the temperature of the surroundings is $\theta_0$. According to Newton's law of cooling,which of the following is the correct graph between $\log_e(\theta - \theta_0)$ and $t$?

$A$ body cools from $75 \, ^\circ \text{C}$ to $65 \, ^\circ \text{C}$ in $2 \, \text{min}$ in a room at $30 \, ^\circ \text{C}$. How long will it take to cool from $55 \, ^\circ \text{C}$ to $45 \, ^\circ \text{C}$ in the same room?

$A$ liquid in a beaker has temperature $\theta(t)$ at time $t$ and $\theta_0$ is the temperature of the surroundings. According to Newton's law of cooling,which of the following is the correct graph between $\log_e(\theta - \theta_0)$ and $t$?

$A$ 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 $39.9^{\circ}C$ (in $, s$)? The temperature of the surroundings is $30^{\circ}C$. Assume Newton's Law of Cooling applies.

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