(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 10-2.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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Class 11

Physics Questions

Chapter

10-2.Heat Transfer

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Newton's Law of Cooling

$A$ liquid is filled in a vessel which is kept in a room with a temperature of $20^{\circ}C$. When the temperature of the liquid is $80^{\circ}C$,it loses heat at the rate of $60 \; cal/sec$. What will be the rate of loss of heat when the temperature of the liquid is $40^{\circ}C$ in $cal/sec$?

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$A$ body takes $10$ minutes to cool down from $62^{\circ}C$ to $50^{\circ}C$. If the temperature of the surrounding is $26^{\circ}C$,then in the next $10$ minutes,the temperature of the body will be ......... $^{\circ}C$.

Medium

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

DifficultJEE MAIN 2014

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On what does the proportionality constant depend in Newton's law of cooling?

Easy

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$A$ solid copper cube of edges $1\;cm$ is suspended in an evacuated enclosure. Its temperature is found to fall from $100^{\circ}C$ to $99^{\circ}C$ in $100\;s$. Another solid copper cube of edges $2\;cm$,with similar surface nature,is suspended in a similar manner. The time required for this cube to cool from $100^{\circ}C$ to $99^{\circ}C$ will be approximately ...... $s$.

Medium

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$A$ liquid is filled in a vessel which is kept in a room with a temperature of $20^{\circ}C$. When the temperature of the liquid is $80^{\circ}C$,it loses heat at the rate of $60 \; cal/sec$. What will be the rate of loss of heat when the temperature of the liquid is $40^{\circ}C$ in $cal/sec$?

$A$ body takes $10$ minutes to cool down from $62^{\circ}C$ to $50^{\circ}C$. If the temperature of the surrounding is $26^{\circ}C$,then in the next $10$ minutes,the temperature of the body will be ......... $^{\circ}C$.

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

On what does the proportionality constant depend in Newton's law of cooling?

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