Let T=the temperature (in °F ) recorded by the thermometer at time t.
The equation governing T is obtained from Newton’s Law of Cooling:
dtdT=k(T−10) where k is some constant.
We are given that T(0)=70 and T(1/2)=50.
T−10dT=kdt Intergrate
∫T−10dT=∫kdt
ln(T−10)=kt+lnC
T−10=Cekt
T=Cekt+10
T(0)=C+10=70=>C=60
T(1/2)=60ek(1/2)+10=50
ek(1/2)=2/3
k=2ln(32)
T(t)=60(94)t+10
a)
T(1)=60(94)1+10=3110(°F)≈36.67(°F)
b)
60(94)t+10=15
(94)t=121
t=ln(4/9)ln(1/12) min≈3.06 min
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