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Thermal Properties of Matter
Newton's Law of Cooling
4 questions, each with its answer.
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A cup of coffee cools from $90^\circ C$ to $80^\circ C$ in $t$ minutes, when the room temperature is $20^\circ C$. The time taken by a similar cup of coffee to cool from $80^\circ C$ to $60^\circ C$ at a room temperature same at $20^\circ C$ is:Newton's law of cooling (average form): $\frac{T_1 - T_2}{t} = K\left[\frac{T_1 + T_2}{2} - T_0\right]$ First cup: $\frac{90 - 80}{t} = K\left[\frac{90 + 80}{2} - 20\right] = 65K$ ...(1) Second cup: $\frac{80 - 60}{t'} = K\left[\frac{80 + 60}{2} - 20\right] = 50K$ ...(2) Dividing (1) by (2): $\frac{t'}{2t} = \frac{65}{50} \Rightarrow t' = \frac{13}{5}t$
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An object kept in a large room having air temperature of $25^\circ C$ takes 12 minutes to cool from $80^\circ C$ to $70^\circ C$. The time taken to cool for the same object from $70^\circ C$ to $60^\circ C$ would be nearly:$\frac{T_1-T_2}{t}=K\left(\frac{T_1+T_2}{2}-T_s\right)$. First: $\frac{10}{12}=K(75-25) \Rightarrow K=\frac{1}{60}$. Second: $\frac{10}{t}=\frac{1}{60}(65-25) \Rightarrow t=15$ min.
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A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be:
Use the average-temperature form of Newton's law twice.
Newton's law of cooling: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T\right)$
$\frac{3T - 2T}{10} = k\left(\frac{5T}{2} - T\right) \Rightarrow \frac{T}{10} = k\left(\frac{3T}{2}\right)$ ...(i)
$\frac{2T - T'}{10} = k\left(\frac{2T + T'}{2} - T\right) = k\left(\frac{T'}{2}\right)$ ...(ii)
Solving (i) and (ii): $T' = \frac{3}{2}T$ -
Certain quantity of water cools from $70^\circ C$ to $60^\circ C$ in the first 5 minutes and to $54^\circ C$ in the next 5 minutes. The temperature of the surroundings is:Newton's law of cooling: $\frac{\theta_1 - \theta_2}{t} = k\left[\frac{\theta_1 + \theta_2}{2} - \theta_0\right]$
$\frac{70 - 60}{5} = k[65 - \theta_0] \Rightarrow 2 = k[65 - \theta_0]$ ...(i)
$\frac{60 - 54}{5} = k[57 - \theta_0] \Rightarrow \frac{6}{5} = k[57 - \theta_0]$ ...(ii)
Dividing (i) by (ii): $\frac{10}{6} = \frac{65 - \theta_0}{57 - \theta_0} \Rightarrow \theta_0 = 45^\circ C$
That is every question in Newton's Law of Cooling.
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