Question

A body cools from temperature \[(3\theta)^{\circ}C\] to \[(2\theta)^{\circ}C\] in 10 minute. Then it cools from \[(2\theta)^{\circ}C\] to \[(\theta_1)^{\circ}C\] in next 10 minutes. The room temperature is \[(\theta)^{\circ}C.\] Assuming that the newton's law of cooling is applicable the value of \[(\theta_1)^{\circ}C\] is ______.

पर्याय

• \[\theta\]

• \[\frac{5}{4}\theta\]

• \[\frac{4}{3}\theta\]

• \[\frac{3}{2}\theta\]

Answer 1

A body cools from temperature \[(3\theta)^{\circ}C\] to \[(2\theta)^{\circ}C\] in 10 minute. Then it cools from \[(2\theta)^{\circ}C\] to \[(\theta_1)^{\circ}C\] in next 10 minutes. The room temperature is \[(\theta)^{\circ}C.\] Assuming that the newton's law of cooling is applicable the value of \[(\theta_1)^{\circ}C\] is \[\frac{3}{2}\theta\].

Explanation:

According to Newton's law of cooling,

\[\frac{\theta_1-\theta_2}{\mathrm{t}}=\mathrm{K}\left[\frac{\theta_1+\theta_2}{2}-\theta_0\right]\]

where, \[\theta_0\] = temperature of surrounding

\[\therefore\quad\frac{3\theta-2\theta}{10}=\mathrm{K}\left[\frac{3\theta+2\theta}{2}-\theta\right]\]

\[\frac{\theta}{10}=\mathrm{K}\times\frac{3\theta}{2}\]

\[\mathrm{K}=\frac{1}{15}\]            ...(i)

After another 10 min, the temperature will be \[\theta_1\]

\[\therefore\quad\frac{2\theta-\theta_1}{10}=\frac{1}{15}\left[\frac{2\theta+\theta_1}{2}-\theta\right]\]

\[\therefore\quad2\theta-\theta_1=\frac{10\theta_1}{30}\]

\[\therefore\quad\theta_1=\frac{3}{2}\theta\]