Question

A bucket full of hot water is kept in a room. It cools from ${75}^{o}C$ to ${70}^{o}C$ in $T_1$ minutes, from ${70}^{o}C$ to ${65}^{o}C$ in $T_2$ minutes and from ${65}^{o}C$ to ${60}^{o}C$ in $T_3$ minutes. Which of the following relation is correct?

• A. $T_1>T_2>T_3$
• B. $T_1<T_2<T_3$
• C. $T_1<T_2>T_3$
• D. $T_1=T_2=T_3$

Answer 1

The correct option is B $T_1<T_2<T_3$

By Newton's law of cooling
$\frac{d{\theta }_2}{{\theta }_2-{\theta }_1}=kdt$
where $k$ is inversely proportional to product of mass and specific heat of the water, ${\theta }_1$ and ${\theta }_2$ are surrounding and average water
temperatures and $dt$ is the time interval.
given that for case (i) $d{\theta }_2={75}^{o}C-{70}^{o}C=5^{o}C$, ${\theta }_2=\frac{75+70}{2}={72.5}^{o}C$, $dt=T_1$
$\frac{d{\theta }_2}{{\theta }_2-{\theta }_1}=kdt\frac{5}{72.5-{\theta }_1}=k{T}_1{T}_1=\frac{5}{k(72.5-{\theta }_1)}$
case (ii) $d{\theta }_2={70}^{o}C-{65}^{o}C=5^{o}C$, ${\theta }_2=\frac{70+65}{2}={67.5}^{o}C$, $dt=T_2$
$\frac{d{\theta }_2}{{\theta }_2-{\theta }_1}=kdt\frac{5}{67.5-{\theta }_1}=k{T}_2{T}_2=\frac{5}{k(67.5-{\theta }_1)}$
case (iii)$d{\theta }_2={65}^{o}C-{60}^{o}C=5^{o}C$, ${\theta }_2=\frac{65+60}{2}={62.5}^{o}C$, $dt=T_3$
$\frac{d{\theta }_2}{{\theta }_2-{\theta }_1}=kdt\frac{5}{62.5-{\theta }_1}=k{T}_3{T}_3=\frac{5}{k(62.5-{\theta }_1)}$
by comparing above three cases $T_1<T_2<T_3$
option (B) is the correct answer