Question 730: Thermodynamics & KTG

Question

A bucket full of hot water is kept in a room. If it cools from $75^{\circ} \mathrm{C}$ to $70^{\circ} \mathrm{C}$ in $t_1$ minutes, from $70^{\circ} \mathrm{C}$ to $65^{\circ} \mathrm{C}$ in $\mathrm{t}_2$ minutes and $65^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ in $t_3$ minutes, then

Accepted Answer

According to Newton's law of cooling, the rate at which an object cools is proportional to the difference between its average temperature and the ambient temperature. Thus, the rate of cooling can be expressed as: $ ext{Rate of cooling} \propto ext{Mean temperature difference} $ This relationship can be mathematically represented as: $ \frac{ ext{Fall in temperature}}{ ext{Time} (t)} \propto \left(\frac{ heta_1 + heta_2}{2} - heta_0 ight) $ Where: $ heta_1$ and $ heta_2$ are the initial and final temperatures, respectively. $ heta_0$ is the ambient room temperature. Let's evaluate each cooling interval: Case 1: From $75^\circ ext{C}$ to $70^\circ ext{C}$ $ ext{Mean temperature} = \left(\frac{75 + 70}{2} ight) = 72.5 $ Case 2: From $70^\circ ext{C}$ to $65^\circ ext{C}$ $ ext{Mean temperature} = \left(\frac{70 + 65}{2} ight) = 67.5 $ Case 3: From $65^\circ ext{C}$ to $60^\circ ext{C}$ $ ext{Mean temperature} = \left(\frac{65 + 60}{2} ight) = 62.5 $ From these calculations, we observe: $ 72.5 > 67.5 > 62.5 $ Since the mean temperature decreases in each subsequent case, the cooling rate is higher initially and decreases over time. This implies: $ t_1 < t_2 < t_3 $

Answer

$\mathrm{t}_1<\mathrm{t}_2<\mathrm{t}_3$

Answer

$\mathrm{t_1>t_2>t_3}$

Answer

$\mathrm{t}_1=\mathrm{t}_2=\mathrm{t}_3$

Answer

$\mathrm{t}_1<\mathrm{t}_2=\mathrm{t}_3$

Thermodynamics & KTG

A bucket full of hot water is kept in a room. If it cools from 75^{\circ} \mathrm{C} to 70^{\circ} \mathrm{C} in t_1 minutes, from 70^{\circ} \mathrm{C} to 65^{\circ} \mathrm{C} in \mathrm{t}_2 minutes and 65^{\circ} \mathrm{C} to 60^{\circ} \mathrm{C} in t_3 minutes, then

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