Question #346468

A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. After one-half minute the thermometer reads 50° F. What is the reading of the thermometer at t  1 min? How long will it take for the thermometer to reach 15° F?


1
Expert's answer
2022-05-31T13:55:11-0400

Let T=T=the temperature (in °F\degree F ) recorded by the thermometer at time t.t.

The equation governing TT is obtained from Newton’s Law of Cooling:


dTdt=k(T10)\dfrac{dT}{dt}=k(T-10)

where kk is some constant.

We are given that T(0)=70T(0) = 70 and T(1/2)=50.T( 1/ 2 ) = 50.


dTT10=kdt\dfrac{dT}{T-10}=kdt

Intergrate


dTT10=kdt\int \dfrac{dT}{T-10}=\int kdt

ln(T10)=kt+lnC\ln(T-10)=kt+\ln C

T10=CektT-10=Ce^{kt}

T=Cekt+10T=Ce^{kt}+10

T(0)=C+10=70=>C=60T(0)=C+10=70=>C=60

T(1/2)=60ek(1/2)+10=50T(1/2)=60e^{k(1/2)}+10=50

ek(1/2)=2/3e^{k(1/2)}=2/3


k=2ln(23)k=2\ln(\dfrac{2}{3})

T(t)=60(49)t+10T(t)=60(\dfrac{4}{9})^t+10

a)


T(1)=60(49)1+10=1103(°F)36.67(°F)T(1)=60(\dfrac{4}{9})^1+10=\dfrac{110}{3}(\degree F)\approx36.67(\degree F)

b)


60(49)t+10=1560(\dfrac{4}{9})^t+10=15

(49)t=112(\dfrac{4}{9})^t=\dfrac{1}{12}

t=ln(1/12)ln(4/9) min3.06 mint=\dfrac{\ln(1/12)}{\ln(4/9)}\ min\approx3.06\ min


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