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Sagot :
Answer:
PLZ GIVE BRAINLESISET
Calculus Newton's Law of Cooling
Sources: #1 – 5: Smith/Minton Calculus 4th ed. #6 – 8: Thomas/Finney Calculus 9th ed.
Use Newton's Law of Cooling T −TS = (TO −TS )e−kt
( ) to solve the following. Round temperature
answers to the nearest tenth of a degree, and time (duration) answers to the nearest hundredth of a
minute.
1) A cup of fast-food coffee is 180°F when freshly poured. After 2 minutes in a room at 70°F, the coffee has
cooled to 165°F. Find the time that it will take for the coffee to cool to 120°F.
165− 70 = (180 − 70)e−2 k
95 =110e−2 k
e−2 k = 95
110
k =
ln
95
110
"
#
$ %
&
'
−2
120 − 70 = (180 − 70)e
−
ln 95
110
"
#
$ %
&
'
−2
"
#
$
$
$
$
%
&
'
'
'
'
t
50 =110e
ln 95
110
!
"
# $
%
&
2
!
"
#
#
#
#
$
%
&
&
&
&
t
e
ln 95
110
!
"
# $
%
&
2
!
"
#
#
#
#
$
%
&
&
&
&
t
= 50
110
t =
ln
50
110
!
"
# $
%
&
ln
95
110
!
"
# $
%
&
2
!
"
#
#
#
#
$
%
&
&
&
&
≈10.76 minutes
2) A bowl of porridge at 200°F (too hot) is placed in a 70°F room. One minute later the porridge has cooled to
180°F. When will the temperature be 120°F (just right)?
180 − 70 = (200 − 70)e−k
110 =130e−k
e−k = 11
13
k = −ln
11
13
"
#
$ %
&
'
120 − 70 = (200 − 70)e
ln 11
13
!
"
# $
%
&t
50 =130e
ln 11
13
!
"
# $
%
&t
e
ln 11
13
!
"
# $
%
&t
= 50
130
t =
ln
5
13
!
"
# $
%
&
ln
11
13
!
"
# $
%
&
≈ 5.72 minutes
3) A smaller bowl of porridge served at 200°F cools to 160°F in one minute. What temperature (too cold) will
this porridge be when the bowl of exercise 2 has reached 120°F (just right)?
160 − 70 = (200 − 70)e−k
90 =130e−k
e−k = 9
13
k = −ln
9
13
"
#
$ %
&
'
T − 70 = (200 − 70)e
ln 9
13
"
#
$ %
&
'(5.72)
T =130e
ln 9
13
"
#
$ %
&
'(5,72)
+ 70
T ≈ 85.9°
4) A cold drink is poured out at 50°F. After 2 minutes of sitting in a 70°F room, its temperature has risen to
56°F.
A) What will the temperature be after 10 minutes?
B) When will the drink have warmed to 66°F?
A)
56 − 70 = (50 − 70)e−2 k
−14 = −20e−2 k
e−2 k = 7
10
k =
ln
7
10
"
#
$ %
&
'
−2
T − 70 = (50 − 70)e
ln 7
10
!
"
# $
%
&
2 (10)
T = 70 +(−20)e
ln 7
10
!
"
# $
%
&
2 (10)
≈ 66.6°F
B)
66 − 70 = (50 − 70)e
ln(0.7)
2 t
−4 = −20e
ln(0.7)
2 t
e
ln(0.7)
2 t
= 4
20
t = ln(0.2)
ln(0.7)
2
≈ 9.02 minutes after it is poured.
5) At 10:07 pm, you find a secret agent murdered. Next to him is a martini that got shaken before he could stir
it. Room temperature is 70°F. The martini warms from 60°F to 61°F in the 2 minutes from 10:07 pm to
10:09 pm. If the secret agent's martinis are always served at 40°F, what was the time of death (rounded to the
nearest minute)?
61− 70 = (60 − 70)e−2 k
−9 = −10e−2 k
e−2 k = 9
10
k = ln(0.9)
−2
60 − 70 = (40 − 70)e
ln(0.9)
2 t
−10 = −30e
ln(0.9)
2 t
e
ln(0.9)
2 t
= 1
3
t =
ln
1
3
!
"
# $
%
&
ln(0.9)
2
≈ 20.85 minutes
The agent was murdered
at approx. 9:46 pm.
6) A hard-boiled egg at 98°C is put into a sink of 18°C water. After 5 minutes, the egg's temperature is 38°C.
Assuming that the surrounding water has not warmed appreciably, how much longer will it take the egg to
reach 20°C?
38−18 = (98−18)e−5k
20 = 80e−5k
e−5k = 20
80
k = ln(0.25)
−5
20 −18 = (38−18)e
ln(0.25)
5 t
2 = 20e
ln(0.25)
5 t
e
ln(0.25)
5 t
= 2
20
t = ln(0.1)
ln(0.25)
5
≈ 8.30 minutes
7) Suppose that a cup of soup cooled from 90°C to 60°C after 10 minutes in a room whose temperature
was 20°C.
A) How much longer would it take the soup to cool to 35°C?
60 − 20 = (90 − 20)e−10 k
40 = 70e−10 k
e−10 k = 40
70
k =
ln
4
7
!
"
# $
%
&
−10
35− 20 = (90 − 20)e
ln 4
7
!
"
# $
%
&
10 t
15 = 70e
ln 4
7
!
"
# $
%
&
10 t
e
ln 4
7
!
"
# $
%
&
10 t
= 15
70
t =
ln
3
14
!
"
# $
%
&
ln
4
7
!
"
# $
%
&
10
≈ 27.53 minutes
B) Instead of being left to stand in the room, the cup of 90°C soup is placed in a freezer whose temperature
is -15°C, and it took 5 minutes to cool to 60°C. How long will it take the soup to cool from 90°C
to 35°C?
60 −(−15) = (90 −(−15))e−5k
75 =105e−5k
e−5k = 75
105
k =
ln
5
7
!
"
# $
%
&
−5
35−(−15) = (90 −(−15))e
ln 5
7
!
"
# $
%
&
5 t
50 =105e
ln 5
7
!
"
# $
%
&
5 t
e
ln 5
7
!
"
# $
%
&
5 t
= 50
105
t =
ln
10
21
!
"
# $
%
&
ln
5
7
!
"
# $
%
&
5
≈11.03 minutes
8) A pan of warm water (46°C) was put into a refrigerator. Ten minutes later, the water's temperature was 39°C.
10 minutes after that it was 33°C. Use Newton's Law of Cooling to estimate the temperature of the
refrigerator.
39 −TS = (46 −TS )e−10 k ⇒ e−10 k = 39 −TS
46 −TS
33−TS = (39 −TS )e−10 k ⇒ e−10 k = 33−TS
39 −TS
39 −TS
46 −TS
= 33−TS
39 −TS
(39 −TS )
2
= (33−TS )(46 −TS )
1521− 78TS +TS
2 =1518− 79TS +TS
2
TS = −3°C
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