Answer to Question #223873 in Chemical Engineering for Lokika

Question

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?

Answer 1

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.

$T_a=30°C\\ T_b=80°C \\ T_f=60°C\\ t=10min\\ (60–30)=(80–60) exp^{-10k}\\( exp^{-10k}=3/2)$

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

$(T-30)=(60-T)exp^{-20k}\\ [exp^{-10k}]²=\dfrac{(T-30)}{(60-T)}\\ \dfrac{(T-30)}{(60-T)}=9/4 \\ \dfrac{30}{(60-T)}=13/4\\\dfrac{(60-T)}{30}=4/13\\ \dfrac{T}{30}=22/13\\ T=30×\dfrac{22}{13}≈50.77°C.$