Question #290004

A thermometer is taken from a room where the temperature is 20°C to the outdoors,

where the temperature is 5°C. After one minute the thermometer read 12°C.Use

Newton’s Law of Cooling to answer the following questions:

a) What will the reading on the thermometer be after one more minute?

b) When will the thermometer read 6°C?


1
Expert's answer
2022-01-24T16:18:36-0500

As per Newton's Law of Cooling, the rate of the cooling is always proportional to the temperature difference between the object and its surroundings.

dTdt=k(T0Ts)\frac{dT}{dt}=-k(T_0-T_s)

Tt=Ts+(T0Ts)ektT_t=T_s+(T_0-T_s)e^{-kt}

T0=20°CT_0=20°C And Ts=5°CT_s=5°C

At t=60, T60=12°CT_{60}=12°C

12=5+(205)e60k12=5+(20-5)e^{-60k}

125=15e60k12-5=15e^{-60k}

715=e60k\frac{7}{15}=e^{-60k}

k=ln(715)60k=-\frac{ln(\frac{7}{15})}{60}

k=0.0127023342k=0.0127023342

a)

After another one minute, t=120s\therefore t=120s

Tt=Ts+(T0Ts)ektT_t=T_s+(T_0-T_s)e^{-kt}

T120=5+(15)e0.0127023342120T_{120}=5+(15)e^{-0.0127023342*120}

=5+0.01611786424=5+0.01611786424

=5.01611786424°C=5.01611786424°C


b)

When temperature will read 6°C

Tt=Ts+(T0Ts)ektT_t=T_s+(T_0-T_s)e^{-kt}

6=5+(15)e0.0127023342t6=5+(15)e^{-0.0127023342t}

1=(15)e0.0127023342t1=(15)e^{-0.0127023342t}

e0.0127023342t=115=0.06666666667e^{-0.0127023342t}=\frac{1}{15}=0.06666666667

t=ln0.066666666670.0127023342t=-\frac{ln 0.06666666667}{0.0127023342}

t=213.19s

=3.55minutes



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