Heating and
Cooling Applications
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Example: A cold drink taken from a
refrigerator has an initial temperature of
5o C. After sitting 25 minutes on the
counter at room temperature of 20o
C its temperature has increased to 10o
C. a. Find the temperature of the drink after 60
minutes. b. Find the time when the drink reaches a temperature
of 18o C. dT/dt = k ( T
˗ TS ) (1) where T(t) is the temperature of the object at
any time t TS is the constant temperature of the
surroundings k is
the proportionality constant To simplify and change the
differential equation into a standard form,
let y(t) = T(t) ˗ TS . The resulting differential
equation takes the form: dy/dt = k y
(2) where y(t)
represents the temperature change,
T(t) ˗ TS |
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Strategy: Assume
dy/dt = ky where y
= T(t) ˗ TS So the temperature difference,
y(t), can be expressed y(t) = y(0)
e kt |
|
Application: y(0) =
5 ˗ 20 = ˗ 15 So
y(t) = ˗ 15 ekt Now y(25)
= 10 ˗ 20 = ˗ 10
= ˗ 15 e25k or e25k = 2/3
and 25 k = ln(2/3) So k
= (1/25) ln(2/3) and y(t)
= ˗ 15 e(1/25) ln(2/3) t = ˗
15 [ (2/3)t/25 ] |
|
a. If t
= 60 minutes, then y(60) = ˗ 15 [ (2/3)60/25
] =
˗ 5.668 = T(60) ˗ 20 So T(60)
= 14.33o C (result) b. 18 ˗
20 = = ˗ 15 [ (2/3)t*/25 ] So
2/15 = [
(2/3)t*/25 ] and ln(2/15) = ( t*/25)
ln (2/3) t*
= 25 [ ln(2/15) / ln (2/3) ] = 124.2 min (result) |
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