Question

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. 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

Answer 1

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.)

Answer 2

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

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

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.