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# Answer to Question #223873 in Chemical Engineering for Lokika

Question #223873

State Newton s law of cooling. If the temperature of the air is 30^{0}C and thetemperature of the body drops from 80°C to 60^{0}C in 10 minutes. What will beits temperature after 20 minutes?

1
2021-08-27T02:07:23-0400

Newton's law of cooling states that the rate at which an object cools is proportional to the difference in temperature between the object and the object's surroundings.

"T_a=30\u00b0C\\\\\n T_b=80\u00b0C \\\\\nT_f=60\u00b0C\\\\\nt=10min\\\\\n(60\u201330)=(80\u201360) exp^{-10k}\\\\( exp^{-10k}=3\/2)"

Let T be the temperature of the body 20 minutes thereafter.

"(T-30)=(60-T)exp^{-20k}\\\\ [exp^{-10k}]\u00b2=\\dfrac{(T-30)}{(60-T)}\\\\ \\dfrac{(T-30)}{(60-T)}=9\/4 \\\\\n\\dfrac{30}{(60-T)}=13\/4\\\\\\dfrac{(60-T)}{30}=4\/13\\\\ \\dfrac{T}{30}=22\/13\\\\ \nT=30\u00d7\\dfrac{22}{13}\u224850.77\u00b0C."

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