1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
Bobcat Park is a rectangular park with an area of 5 2/5 square miles.Its width is 1 11/25 miles.How long is the park?
ELEN [110]
I believe it is 3.75 miles. 5 2/5=5.4 and 1 11/25=1.44. 5.4/1.44=3.75.
4 0
3 years ago
Which number is 5 more than 2? Which number is 4 less than -3? PLS HELP ME ILL GIVE YOU BRAINLIST
vaieri [72.5K]

5+2=7 4- -3= -7 hdkb sus heyitsme

6 0
3 years ago
Read 2 more answers
Please please help me on this question
Delvig [45]
-7x^2 - 6x^2 + 3x + 5x - 5 + 2
-13x^2 + 8x - 3
5 0
3 years ago
Plss help if can!! Thanks!
Novosadov [1.4K]
The first one is 2 the second one is 3
5 0
3 years ago
A blue whale calf weighed 2725 kilograms at birth. A blue whale calf gains 90 kilograms of weight each day for the first 240 day
Ivanshal [37]

Answer:the weight after 225 days is

22885 kilograms

Step-by-step explanation:

The initial weight of the blue whale calf at birth is 2725 kilograms. blue whale calf gains 90 kilograms of weight each day for the first 240 days after its birth. The weight increases in arithmetic progression. This means that the first term of the sequence, a is 2725, the common difference, d is 90.

The formula for the nth term of an arithmetic sequence is expressed as

Tn = a + (n - 1)d

Where

n is the number of terms of the sequence.

a is the first term

d is the common difference

We want to determine its weight, T225 after 225 days after it’s birth. It means that n = 225

Therefore

T225 = 2725 + (225 - 1)90

T225 = 2725 + 224×90 = 2725 + 20160

T225 = 22885

8 0
3 years ago
Other questions:
  • A salesperson receives a 3% commission on sales. The salesperson receives $180 in commission. What is the amount of sales?
    15·2 answers
  • Please help I will give brainliest
    10·1 answer
  • How to answer this qestion?
    14·1 answer
  • Rachel, Kim, Lori each measure the length of a rope. Rachel says the rope is 15 feet long . Kim day it’s 180 inches long . Lori
    12·1 answer
  • How much greater is the sum of the first 100 triangular numbers than the sum of the first 99 triangular numbers? Answer: It is b
    9·2 answers
  • Solve each inequality.
    11·1 answer
  • You store square notepaper in a cube shape box with an inside edge length of 3 inches. What is the volume of the box
    5·1 answer
  • Given f(x)=x^3 and g(x)= 1-5x^2
    5·1 answer
  • Solve the equation y + 5=-12 <br> A y=-19<br> B y=-17<br> C y=-7<br> D y=7
    5·2 answers
  • The number of books read in each week of a month are 4, 6, 8, and 5. What<br> is the range?
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!