You place a cup of 200 degrees F coffee on a table in a room that is 67 degrees F, and 10 minutes later, it is 195 degrees F. Approximately how long will it be before the coffee is 180 degrees F? Use Newton's law of cooling: T (t) = T[a] + (T[o]-T[a]) e^-kt A. 40 minutes B. 15 minutes C. 1 hour D. 35 minutes

Question

Mathematics

Aedan

30 May, 05:01

You place a cup of 200 degrees F coffee on a table in a room that is 67 degrees F, and 10 minutes later, it is 195 degrees F. Approximately how long will it be before the coffee is 180 degrees F? Use Newton's law of cooling: T (t) = T[a] + (T[o]-T[a]) e^-kt A. 40 minutes B. 15 minutes C. 1 hour D. 35 minutes

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Answers (2)

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Addison Stevenson · Answer 1

43 mins to keep it

short but other than that same as the the other guy

Erin Barron · Answer 2

195=67 + (200-67) e^ (-10k)

195=67+133e^ (-10k)

128=133e^ (-10k)

128/133=e^ (-10k)

ln (128/133) = - 10k

k=-ln (128/133) / 10

T (t) = 67+133e^ (tln (128/133) / 10), if T (t) = 180

180=67+133e^ (tln (128/133) / 10)

113=133e^ (tln (128/133) / 10)

113/133=e^ (tln (128/133) / 10)

ln (113/133) = tln (128/133) / 10

10ln (113/133) = tln (128/133)

t=10ln (113/133) / ln (128/133)

t≈42.5 minutes ...

t≈40 minutes (to nearest 5 minutes? : P)