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

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.

Answer 2

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.)