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Q. An object is cooled from $ 75{}^\circ C $ to $ 65{}^\circ C $ in 2 minute in a room at $ 30{}^\circ C $ . The time taken to cool another object form $ 55{}^\circ C $ to $ 45{}^\circ C $ in the same room in minute is:

EAMCETEAMCET 1996

Solution:

Accord to Newtons law of cooling Rate of cooling $ \propto $ excess of temperature or rate of cooling $ =K\times $ excess of temperature. If $ {{\theta }_{1}} $ is initial temperature, $ {{\theta }_{2}} $ is the final temperature and $ {{\theta }_{0}} $ is the room temperature, then $ \frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=K\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right) $ Since, body cools from, $ {{75}^{o}} $ to $ {{65}^{o}} $ in 2 min and surrounding temperature is $ {{30}^{o}}. $ $ \frac{75-65}{2}=K\left( \frac{75+65}{2}-30 \right) $ or $ 5=K\times 40\Rightarrow K=\frac{5}{40}=\frac{1}{8} $ When body cools from $ 55{{\,}^{o}}C $ to $ 45{{\,}^{o}}C, $ we have $ \frac{55-45}{{{t}_{2}}}=\frac{1}{8}\left( \frac{55+45}{2}-30 \right) $ $ \frac{10}{{{t}_{2}}}=\frac{1}{8}\times 20 $ or $ {{t}_{2}}=\frac{10\times 8}{20}=4\min $