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A thermometer is taken from an inside room to the outside, where the air temperature is 20° f. after 1 minute the thermometer reads 70° f, and after 5 minutes it reads 45° f. what is the initial temperature of the inside room

Respuesta :

DeanR

The change in the temperature is proportional to the difference in temperatures. Let's call the initial temperature of the room and the thermometer [tex]E_0[/tex] and the current temperature [tex]E(t)[/tex] where [tex]t[/tex] is the elapsed time in minutes.  The outside temperature is 20 F.

Newton's Law of Cooling:

[tex]E(t) - 20 = (E_0 - 20) e^{- k t}[/tex]

We're given [tex]E(1)=70 \textrm{ and } E(5)=45[/tex]

[tex]70-20 = (E_0 - 20) e^{-k}[/tex]

[tex]45-20 =(E_0 - 20) e^{-5k}[/tex]

Dividing,

[tex]\dfrac{50}{25}=e^{4k}[/tex]

[tex]k = \dfrac 1 4 \ln 2 [/tex]

Now,

[tex]E(t) - 20 = (E_0 - 20) e^{- k t}[/tex]

[tex]E_0 = 20 + (E(t) - 20)e^{k t}[/tex]

[tex]E_0 = 20 + (E(1) - 20)e^{k}[/tex]

[tex]E_0 = 20 + (70 - 20)e^{\ln 2/4}[/tex]

[tex]E_0 = 20 + 50 \cdot 2^{1/4} \approx 79.46 \textrm{ degrees F} [/tex]


fichoh

Using the Newton's law of cooling principle, the initial temperature of the inside room would be 79.44°F

According to the Newton's Law of Cooling :

  • [tex]T(t) = T_{s} + (T_{o} - T_{s})e^{-kt} [/tex]

  • Temperature at a given time, t = 5 minutes = 45°

The equation can be expressed thus :

[tex]\deltaT (T_{o} - T_{s})e^{-kt} [/tex]

[tex]70 - 20 = (T_{o} - 20)e^{-k} [/tex]

[tex]70 - 45 = (T_{o} - 20)e^{-5k} [/tex]

[tex] 50 = (T_{o} - 20)e^{-k} [/tex] - - - (1)

[tex] 25 = (T_{o} - 20)e^{-5k} [/tex] - - - - (2)

Divide (1) and (2) - - -

[tex]\frac{50}{25} = e^{-k + 5k} [/tex]

[tex]\frac{50}{25} = e^{4kt} [/tex]

[tex] 2 = e^{4kt} [/tex]

Take the ln

[tex] In 2 = 4k [/tex]

k = 0.173

The initial temperature is defined thus :

[tex]T(o) = T_{s} + (T_{t} - T_{s})e^{kt} [/tex]

[tex]T(o) = 20 + (70 - 20)e^{0.173(1)} [/tex]

[tex]T(o) = 20 + (50 \times 1.1888) = 79.44°F[/tex]

Therefore, the initial temperature is 79.44°F

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