(C) The rate of cooling $R_C$ is given by the formula: $R_C = \frac{dQ}{dt} \cdot \frac{1}{mc} = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{mc}$.Since

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Rate of cooling of $A$ is twice that of $B$

(C) The rate of cooling $R_C$ is given by the formula: $R_C = \frac{dQ}{dt} \cdot \frac{1}{mc} = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{mc}$.Since $m = \rho V$,where $\rho$ is density and $V$ is volume,we have $R_C = \frac{A \varepsilon \sigma (T^4 - T_0^4)}{V \rho c}$.For a sphere,$A = 4 \pi r^2$ and $V = \frac{4}{3} \pi r^3$,so $\frac{A}{V} = \frac{3}{r}$.Thus,$R_C \propto \frac{1}{r} \propto \frac{1}{\text{Diameter}}$.Given that the diameter of $A$ is half that of $B$ $(D_A = \frac{1}{2} D_B)$,the rate of cooling of $A$ will be twice that of $B$ $(R_{C,A} = 2 R_{C,B})$.

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

Medium

There are two spherical balls $A$ and $B$ of the same material with the same surface,but the diameter of $A$ is half that of $B$. If $A$ and $B$ are heated to the same temperature and then allowed to cool,then

A
Rate of cooling is same in both
B
Rate of cooling of $A$ is four times that of $B$
C
Rate of cooling of $A$ is twice that of $B$
D
Rate of cooling of $A$ is $\frac{1}{4}$ times that of $B$

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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 cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes. If the room temperature is $25^{\circ}C$ and assuming Newton's law of cooling to hold good,the temperature of the body at the end of the next $10$ minutes will be ......... $^{\circ}C$.

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If a piece of metal is heated to temperature $\theta$ and then allowed to cool in a room which is at temperature $\theta_0$,the graph between the temperature $T$ of the metal and time $t$ will be closest to:

MediumJEE MAIN 2013

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Two metallic spheres $P$ and $Q$ are made of the same material and have the same smoothness,but the weight of $P$ is $8$ times that of $Q$. If the two are heated to the same temperature and left to cool,then the ratio of the rate of cooling of $Q$ to that of $P$ is:

MediumAP EAMCET 2019

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$A$ solid sphere and a hollow sphere of the same material and size are heated to the same temperature and allowed to cool in the same surroundings. If the temperature difference between each sphere and its surroundings is $T$,then

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There are two spherical balls $A$ and $B$ of the same material with the same surface,but the diameter of $A$ is half that of $B$. If $A$ and $B$ are heated to the same temperature and then allowed to cool,then

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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 cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes. If the room temperature is $25^{\circ}C$ and assuming Newton's law of cooling to hold good,the temperature of the body at the end of the next $10$ minutes will be ......... $^{\circ}C$.

If a piece of metal is heated to temperature $\theta$ and then allowed to cool in a room which is at temperature $\theta_0$,the graph between the temperature $T$ of the metal and time $t$ will be closest to:

Two metallic spheres $P$ and $Q$ are made of the same material and have the same smoothness,but the weight of $P$ is $8$ times that of $Q$. If the two are heated to the same temperature and left to cool,then the ratio of the rate of cooling of $Q$ to that of $P$ is:

$A$ solid sphere and a hollow sphere of the same material and size are heated to the same temperature and allowed to cool in the same surroundings. If the temperature difference between each sphere and its surroundings is $T$,then

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