Question #229220

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
Expert's answer
2021-08-28T06:15:59-0400

Newton's law is a physical law that states that the rate at which a substance is cooling is proportional to the difference in temperature between the object and the object's surroundings.


Ta=30°CTb=80°CTf=60°Ct=10min(6030)=(8060)exp10k(exp10k=3/2)T_a=30°C\\ T_b=80°C \\ T_f=60°C\\ t=10min\\ (60–30)=(80–60) 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}]²=\dfrac{(T-30)}{(60-T)}\\ \dfrac{(T-30)}{(60-T)}=9/4 \\ T-30 = 9/4(60-T)\\ T-30 = 135 -9/4T\\ T+9/4T = 135+30\\ T× 13/4 = 165\\ T = 165 × 4/13 \\ T≈50.77°C.

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