(B) According to Newton's law of cooling,$\frac{d\theta}{dt} = -k(\theta - \theta_0)$,where $k = \frac{eA\sigma}{ms}$.Integrating this,we get $\log(\t

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(B) According to Newton's law of cooling,$\frac{d\theta}{dt} = -k(\theta - \theta_0)$,where $k = \frac{eA\sigma}{ms}$.Integrating this,we get $\log(\theta - \theta_0) = -kt + C$.The slope of the graph $\log(\theta - \theta_0)$ versus $t$ is $-k$.From the graph,the magnitude of the slope for body $B$ is greater than that for body $A$,so $k_B > k_A$.Since $k = \frac{eA\sigma}{ms}$ and $m, e, A, \sigma$ are the same for both bodies,$k \propto \frac{1}{S}$.Therefore,$k_B > k_A$ implies $S_B S_B$.

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

Medium

Two bodies $A$ and $B$ of equal masses,surface area,and emissivity are cooling under Newton's law of cooling from the same initial temperature. Their cooling curves are represented by the graph. If $\theta$ is the instantaneous temperature of the body and $\theta_0$ is the temperature of the surroundings,then the relationship between their specific heats $S_A$ and $S_B$ is:

A
$S_A = S_B$
B
$S_A > S_B$
C
$S_A < S_B$
D
None of these

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2 Thermal Properties of Matter: Heat Transfer

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Newton's Law of Cooling

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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$A$ body cools from a temperature $3\theta$ to $2\theta$ in $10 \text{ minutes}$. The room temperature is $\theta$. The temperature of the body at the end of the next $10 \text{ minutes}$ is '$x$'. Assuming that Newton's law of cooling is applicable, the value of '$x$' will be

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If the initial temperatures of a metallic sphere and a disc,of the same mass,radius,and material,are equal,then the ratio of their rate of cooling in the same environment will be:

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$A$ body cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes. If the room temperature is $25^{\circ}C$ and Newton's law of cooling is applicable,what will be the temperature of the body after another $10$ minutes in $^{\circ}C$?

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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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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.

$A$ body cools from a temperature $3\theta$ to $2\theta$ in $10 \text{ minutes}$. The room temperature is $\theta$. The temperature of the body at the end of the next $10 \text{ minutes}$ is '$x$'. Assuming that Newton's law of cooling is applicable, the value of '$x$' will be

If the initial temperatures of a metallic sphere and a disc,of the same mass,radius,and material,are equal,then the ratio of their rate of cooling in the same environment will be:

$A$ body cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes. If the room temperature is $25^{\circ}C$ and Newton's law of cooling is applicable,what will be the temperature of the body after another $10$ minutes in $^{\circ}C$?

$A$ 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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