740
Chapter 6 | Applications of Integration
180°F. When is the coffee first cool enough to serve? When is the coffee too cold to serve? Round answers to the nearest half minute.
Solution We have
T = ⎛
⎝ T 0 − T a ⎞
⎠ e −
kt + T
a 180 = (200 − 70) e − k (2) +70 110 = 130 e −2 k 11 13 = e −2 k ln 11 13 = −2 k
ln11−ln13 = −2 k
k = ln13−ln11 2 .
Then, the model is
⎛ ⎝ ln11−ln13 2 ⎞ ⎠ t
T =130 e
+70.
The coffee reaches 175°F when
175 = 130 e ⎛ 105 = 130 e ⎛
⎝ ln11−ln13 2 ⎞ ⎠ t ⎝ ln11−ln13 2 ⎞ ⎠ t
+70
⎛ ⎝ ln11−ln13 2 ⎞ ⎠ t
21 26
= e
ln 21 26 = ln11−ln13 2 t ln21−ln26 = ln11−ln13 2 t t = 2(ln 21 − ln 26) ln11−ln13
≈2.56. The coffee can be served about 2.5 minutes after it is poured. The coffee reaches 155°F at
155 = 130 e ⎛ 85 = 130 e ⎛
⎝ ln11−ln13 2 ⎞ ⎠ t ⎝ ln11−ln13 2 ⎞ ⎠ t
+70
⎛ ⎝ ln11−ln13 2 ⎞ ⎠ t
17 26
= e
⎛ ⎝ ln11−ln13 2 ⎞ ⎠ t t = 2(ln 17 − ln 26) ln11−ln13
ln17−ln26 =
≈5.09.
The coffee is too cold to be served about 5 minutes after it is poured.
6.45 Suppose the room is warmer (75°F) and, after 2 minutes, the coffee has cooled only to 185°F. When is the coffee first cool enough to serve? When is the coffee be too cold to serve? Round answers to the nearest half minute.
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