Question

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

Answer 1

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)

Answer 2

43 mins to keep it
short but other than that same as the the other guy