A can of soda at 80°F is placed in a refrigerator that maintains a constant temperature of 37°F. The temperature T of the soda t minutes after it is placed in the refrigerator is given by
T(t) = 37 + 43e−0.058t.
2 answers
Whats the Question?
Supose you want the soda at 41ºF(5ºC), then:
41=37+43e-0,058t
4=43e-0,058t
e-0,058t=0,093 or 1/e0,058t=0,093
e0,058t = 10,75
0,058t 0 ln 10,75
t=2,375/0,058
Finally t=40,95 min
41=37+43e-0,058t
4=43e-0,058t
e-0,058t=0,093 or 1/e0,058t=0,093
e0,058t = 10,75
0,058t 0 ln 10,75
t=2,375/0,058
Finally t=40,95 min