Question

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 1

`T=51.64^\circ F`

`t=180.10s`

Step-by-step explanation:

The Newton's law in this case is:

`T(t)=T_m+Ce^(kt)`

Here,
`T_m` is the air temperture, C and k are constants.

We have

`70^\circ F` in
`t=0`

So:

`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`

And we have
`60^\circ F` in
`t=30 s`, So:

`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))=(50^\circ F)/(60^\circ F)\\(30)k=ln((50)/(60))\\k=(ln((50)/(60)))/(30)=-0.0061`

Now, we have:

`T=10^\circ F+(60^\circ F)e^(-0.0061t)(1)`

Applying (1) for
`t=1 min=60s`:

`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`

Applying (1) for
`T=30^\circ F`:

`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((20)/(60))\\t=(ln((20)/(60)))/(-0.0061)=180.10s`