Here are my steps:

u(t)=T+(uo-T)e^kt where k<0

300=70+(450-70)e^5k

300=70+(380)e^5k

230=380e^5k

e^5k=23/38

5k lne= 23/38

k=ln 23/38/5

k=-.1004183888

---------

230=70+380e^5k

380e^5k=160

e^kt=8/19

kt=ln 8/19

kt/k=ln 8/19/k

t=8.6 minutes.

8.6 minutes + 5 minutes initial cooling time = 5:13 pm.

Question

19-12-2017
Mathematics

contestada

Here are my steps:

u(t)=T+(uo-T)e^kt where k<0

300=70+(450-70)e^5k

300=70+(380)e^5k

230=380e^5k

e^5k=23/38

5k lne= 23/38

k=ln 23/38/5

k=-.1004183888

---------

230=70+380e^5k

380e^5k=160

e^kt=8/19

kt=ln 8/19

kt/k=ln 8/19/k

t=8.6 minutes.

8.6 minutes + 5 minutes initial cooling time = 5:13 pm.

Respuesta :

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jimthompson5910 jimthompson5910 · Answer 1 · 2017-12-19T02:41:24+01:00

You have the correct approximate value of k. The part after that is where things get a bit strange.

You don't plug in t = 5 again. We don't know the value of t yet. We're trying to solve for it.

We do know u(t) is 135 as we want the temperature to be 135 degrees. Plug this in and isolate t.

So
u(t) = 70+380e^(-0.1004183888t)
135 = 70+380e^(-0.1004183888t)
135-70 = 380e^(-0.1004183888t)
65 = 380e^(-0.1004183888t)
380e^(-0.1004183888t) = 65
e^(-0.1004183888t) = 65/380
e^(-0.1004183888t) = 13/76
-0.1004183888t = ln(13/76)
-0.1004183888t = -1.7657839828248
t = -1.7657839828248/(-0.1004183888)
t = 17.5842692152893

So it takes about 17.58 minutes for the item to cool to the desired temp

Here are my steps: u(t)=T+(uo-T)e^kt where k<0 300=70+(450-70)e^5k 300=70+(380)e^5k 230=380e^5k e^5k=23/38 5k lne= 23/38 k=ln 23/38/5 k=-.1004183888 --------- 230=70+380e^5k 380e^5k=160 e^kt=8/19 kt=ln 8/19 kt/k=ln 8/19/k t=8.6 minutes. 8.6 minutes + 5 minutes initial cooling time = 5:13 pm.

Respuesta :

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