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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 60° F. What is the reading of the thermometer at t = 1 min? (Round your answer to two decimal places.) ° F How long will it take for the thermometer to reach 30° F? (Round your answer to two decimal places.)

Answer :

sebasacevedop

Answer:

[tex]T=51.64^\circ F[/tex]

[tex]t=180.10s[/tex]

Explanation:

The Newton's law in this case is:

[tex]T(t)=T_m+Ce^{kt}[/tex]

Here, [tex]T_m[/tex] is the air temperture, C and k are constants.

We have

[tex]70^\circ F[/tex] in [tex]t=0[/tex]

So:

[tex]T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F[/tex]

And we have [tex]60^\circ F[/tex] in [tex]t=30 s[/tex], So:

[tex]T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061[/tex]

Now, we have:

[tex]T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)[/tex]

Applying (1) for [tex]t=1 min=60s[/tex]:

[tex]T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F[/tex]

Applying (1) for [tex]T=30^\circ F[/tex]:

[tex]30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s[/tex]

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