(C) The rate of cooling is given by $\frac{d\theta}{dt} = \frac{A \epsilon \sigma (T^4 - T_0^4)}{mc}$.Since the temperature change $\Delta \theta$ is

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(C) The rate of cooling is given by $\frac{d\theta}{dt} = \frac{A \epsilon \sigma (T^4 - T_0^4)}{mc}$.Since the temperature change $\Delta \theta$ is small,we can write $\frac{\Delta \theta}{t} \propto \frac{A}{mc}$.Here,$m = \rho V = \rho a^3$ and $A = 6a^2$,where $a$ is the edge length.Thus,$t \propto \frac{mc}{A} \propto \frac{\rho a^3 c}{6a^2} \propto a$.Therefore,$\frac{t_1}{t_2} = \frac{a_1}{a_2}$.Given $t_1 = 100\;s$,$a_1 = 1\;cm$,and $a_2 = 2\;cm$.$\frac{100}{t_2} = \frac{1}{2} \implies t_2 = 200\;s$.

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

Medium

$A$ 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 ...... $s$.

A
$25$
B
$50$
C
$200$
D
$400$

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Class 11

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10-2.Heat Transfer

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Newton's Law of Cooling

$A$ hot body placed in air cools down to a lower temperature. The rate of decrease of temperature is proportional to the temperature difference from the surrounding. The body loses $60 \%$ and $80 \%$ of the maximum heat it can lose in time $t_1$ and $t_2$ respectively. The ratio $t_2 / t_1$ will be

EasyTS EAMCET 2019

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$A$ liquid cools from $50^{\circ}C$ to $45^{\circ}C$ in $5$ minutes and from $45^{\circ}C$ to $41.5^{\circ}C$ in the next $5$ minutes. The temperature of the surrounding is......... $^{\circ}C$.

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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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$A$ body cools down from $52.5^{\circ} C$ to $47.5^{\circ} C$ in $5 \ min$ and from $47.5^{\circ} C$ to $42.5^{\circ} C$ in $7.5 \ min$. Then,the temperature of the surroundings is (in $^{\circ} C$)

MediumAP EAMCET 2021

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Two bodies $A$ and $B$ of same mass,area,and surface finish with specific heats $S_A$ and $S_B$ $(S_A > S_B)$ are allowed to cool for a given temperature range. The temperature varies with time as shown in the graphs. Which graph correctly represents the cooling process?

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$A$ liquid cools from $50^{\circ}C$ to $45^{\circ}C$ in $5$ minutes and from $45^{\circ}C$ to $41.5^{\circ}C$ in the next $5$ minutes. The temperature of the surrounding is......... $^{\circ}C$.

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:

$A$ body cools down from $52.5^{\circ} C$ to $47.5^{\circ} C$ in $5 \ min$ and from $47.5^{\circ} C$ to $42.5^{\circ} C$ in $7.5 \ min$. Then,the temperature of the surroundings is (in $^{\circ} C$)

Two bodies $A$ and $B$ of same mass,area,and surface finish with specific heats $S_A$ and $S_B$ $(S_A > S_B)$ are allowed to cool for a given temperature range. The temperature varies with time as shown in the graphs. Which graph correctly represents the cooling process?

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