Question

Two bottles A and B have radii ${R_A}$ and ${R_B}$and heights ${h_A}$ and ${h_B}$ respectively with​ and ${h_B} = 2{h_A}$​. These are filled with hot water at ${60^0}C$. Consider that heat loss for the bottles takes place only from side surfaces. If the time the water takes to cool down to ${50^0}C$ is ${t_A}$ and ${t_B}$ for bottles A and B, respectively, then ${t_A}$ and ${t_B}$ are best related as.
A) ${t_A} = {t_B}$
B) ${t_B} = 2{t_A}$
C) ${t_B} = 4{t_A}$
D) ${t_B} = \dfrac{t_A}{2}$

Answer 1

First we find the relation between rate of heat transfer $\dfrac{{dQ}}{{dt}}$ and change in temperature $\Delta T$. Then we convert mass in form of density and volume. After that we convert volume into height and height terms.
After that we get how ${t_A}$ and ${t_B}$ are related to each other.

Formula used:
We are using Newton’s law of cooling formula $\dfrac{{dQ}}{{dt}} = hA\Delta T$. Then to calculate the relation between${t_A}$ and , we use $mass = density \times volume$ or in numerically $m = \rho \times V$.

Complete step by step solution:
Given: Radius of two bottles A and B as ${R_A}$and${R_B}$respectively.
Heights of A and B are ${h_A}$ and ${h_B}$respectively.
Relation between ${R_A}$and${R_B}$: ${R_B} = 2{R_A}$
Relation between${h_A}$ and ${h_B}$:${h_B} = 2{h_A}$
According to Newton’s law of cooling
$\dfrac{{dQ}}{{dt}} = hA\Delta T$
$\Rightarrow \Delta Q = hA\Delta T\Delta t$
Ratio between $\Delta Q$of bottle having radius A to bottle having radius B, is given by:
$\dfrac{{\Delta {Q_A}}}{{\Delta {Q_B}}} = \dfrac{{{A_A}{t_A}}}{{{A_B}{t_B}}}$
\Rightarrow \dfrac{{{m_A}s\Delta T}}{{{m_B}s\Delta T}} = \dfrac{{{A_A}{t_A}}}{{{A_B}{t_B}}}$$$$\left[ {\because \Delta Q = ms\Delta T} \right]
$\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{{{A_A}{t_A}}}{{{A_B}{t_B}}}$
\Rightarrow \dfrac{{\rho {V_A}}}{{\rho {V_A}}} = \dfrac{{{A_A}{t_A}}}{{{A_B}{t_B}}}$$$$\left[ {\because m = \rho \times V} \right]
$\dfrac{{{V_A}}}{{{V_A}}} = \dfrac{{{A_A}{t_A}}}{{{A_B}{t_B}}}$
\Rightarrow \dfrac{{{h_A}{A_A}}}{{{h_B}{A_B}}} = \dfrac{{{A_A}{t_A}}}{{{A_B}{t_B}}}$$$$[\because V = hA]

$$\dfrac{{{h_A}}}{{{h_B}}} = \dfrac{{{t_A}}}{{{t_B}}} \\\ \dfrac{{{h_A}}}{{2{h_A}}} = \dfrac{{{t_A}}}{{{t_B}}} \\\$$

$\dfrac{{{t_A}}}{{{t_B}}} = \dfrac{1}{2}$
$\therefore {t_B} = 2{t_A}$

Hence, the correct option is B.

Additional information: The difference in temperatures between the body and its surroundings is explained by Newton's law of cooling. This law describes that through radiation, the rate at which an exposed body changes temperature. If the temperature difference between a body and its surrounding is small, then the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed.

Note: Students must be careful to use laws related to temperature. There are many laws for temperature. In this question only Newton’s law of cooling is used. On solving questions, students must be understood that in question only radius and height are given. So, they convert quantities given in formula in form of radius or height. So, we can find the relation between time taken to cool down to bottles A and B. In other words, the relation between ${t_A}$ and ${t_B}$ can be calculated.