# A hot liquid kept in a beaker cools from 87c to 70c in 2 minutes.if the surrounding temperature is 30c,what is time of cooling of the same liquid from 60c to 50c?

According to Newton's law of cooling, we have

$\\ \mathrm{As}, \frac{d T}{d t}=-K\left[\frac{T_{1}+T_{2}}{2}-T_{0}\right] \\\therefore \frac{80-70}{2 \times 60}=-K\left[\frac{80+70}{2}-30\right] \frac{10}{2 \times 60}=-K \times 45 \\ And \frac{60-50}{t}=-K\left[\frac{60+50}{2}-30\right] \frac{10}{t}=-K \times 25 \ldots(\mathrm{ii}) \\ Dividing (i) by (ii), we get \frac{t}{2 \times 60}=\frac{45}{25}=\frac{9}{5} or t=216 s$

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