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dimulka [17.4K]
3 years ago
9

- The temperature of a hot liquid is 100 degrees. The liquid is placed in a refrigerator

Mathematics
1 answer:
oee [108]3 years ago
5 0

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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