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KengaRu [80]
3 years ago
9

A bottle of white wine at room temperature (68°f) is placed in a refrigerator at 4 p.m. its temperature after t hr is changing a

t the rate of −18e−0.6t °f/hour. by how many degrees will the temperature of the wine have dropped by 5 p.m.? (round your answer to one decimal place.) 17.0 changed: your submitted answer was incorrect. your current answer has not been submitted. your answer has the wrong sign.°f what will the temperature of the wine be at 5 p.m.? (round your answer to one decimal place.) °f
Mathematics
2 answers:
musickatia [10]3 years ago
7 0

Answer:

The temperature of the wine have dropped by 0.9°F.

The temperature of the wine will be 58.1 °F at 5 PM.

Step-by-step explanation:

We have been given that a bottle of white wine at room temperature (68°f) is placed in a refrigerator at 4 p.m. its temperature after t hr is changing at the rate of -18e^{-0.6t} °f/hour.

We will use newton'e law of cooling to solve our given problem.

T(t)=Ce^{-kt}+T_a, where,

T_a = Initial temperature.

Upon substituting T_a=68 and Ce^{-kt}=-18e^{-0.6t}, we will get:

T(t)=-18e^{-0.6t}+68

To find the temperature at 5 p.m. we will substitute t=5 in our function as:

T(5)=-18e^{-0.6(5)}+68

T(5)=-18e^{-3}+68

T(5)=-18(0.0497870683678639)+68

T(5)=-0.8961672306215502+68

T(5)=67.103832769\approx 67.10

Now, we will subtract 67.10 from 68 to find the temperature dropped by 5 PM.

68-67.10=0.9

Therefore, the temperature dropped by 0.9°F.

To find the temperature of wine at 5 PM, we will substitute x=1 as 5 PM is 1 hour later 4 PM.

T(1)=-18e^{-0.6(1)}+68

T(1)=-18e^{-0.6}+68

T(1)=-9.87860944969+68

T(1)=58.12139055031\approx 58.1

Therefore, the temperature of the wine will be 58.1 °F at 5 PM.

Nat2105 [25]3 years ago
5 0
After an hour (4pm to 5pm) the temperature drops 18e⁻⁰˙⁶=9.9℉, so the temperature after an hour will be 68-9.9=58.1℉.
(The change -18e⁻⁰˙⁶ is negative indicating a drop in temperature.)
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