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(A) According to Newton's Law of Cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $-dT

Vedclass · Accepted Answer

$t_1 < t_2 < t_3$

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(A) According to Newton's Law of Cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $-dT/dt = k(T - T_S)$.Here,$T$ is the average temperature of the body and $T_S$ is the temperature of the surroundings.For a small temperature drop $\Delta T = 10 \, ^\circ	ext{C}$,the time taken $t$ is given by $t \propto 1 / (T_{avg} - T_S)$.$1$. For $100 \, ^\circ	ext{C}$ to $90 \, ^\circ	ext{C}$,$T_{avg} = 95 \, ^\circ	ext{C}$,so $t_1 \propto 1 / (95 - T_S)$.$2$. For $90 \, ^\circ	ext{C}$ to $80 \, ^\circ	ext{C}$,$T_{avg} = 85 \, ^\circ	ext{C}$,so $t_2 \propto 1 / (85 - T_S)$.$3$. For $80 \, ^\circ	ext{C}$ to $70 \, ^\circ	ext{C}$,$T_{avg} = 75 \, ^\circ	ext{C}$,so $t_3 \propto 1 / (75 - T_S)$.Since the average temperature $T_{avg}$ decreases in each step,the denominator $(T_{avg} - T_S)$ decreases,which means the time $t$ increases.Therefore,$t_1 < t_2 < t_3$.

The temperature of an iron block falls from $100 \, ^\circ\text{C}$ to $90 \, ^\circ\text{C}$ in time $t_1$,from $90 \, ^\circ\text{C}$ to $80 \, ^\circ\text{C}$ in time $t_2$,and from $80 \, ^\circ\text{C}$ to $70 \, ^\circ\text{C}$ in time $t_3$. Then:

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