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?

Expert's answer

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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