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A hot bowl of soup cools according to Newtons law of cooling. Its temperature (in degrees Fahrenheit) at time t is given by T(t)= 68+144e^-0.04t, where t is given in minutes. Answer these questions What was the initial temperature of the soup? What was the temperature of the soup after 15 minutes? How long after serving is the soup 125 degrees Fahrenheit?

Sagot :

Giovi
212°F
147°F
23.2 min

1) Initial temperature for t=0
T(0)=68+144e^(0)=68+144=212°F
2) After 15 min
T(15)=68+144e^(-0.04*15)=68+144e^(-0.6)=147°F
3) 125°F -> t=?
125=68+144e^(-0.04t)
144e^(-0.04t)=125-68=57
e^(-0.04t)=57/144=0.4 you can apply ln on both sides:
-0.04t=ln(0.4) solving you get t=23.3 min

Initial temperature of the soup is 212°F.

147°C is the the temperature of the soup after 15 minutes.

After 23.2 minutes soup will reach 125 degrees Fahrenheit.

What is temperature?

Temperature is a physical quantity that expresses hot and cold or a measure of the average kinetic energy of the atoms or molecules in the system.

According to questions, temperature (in degrees Fahrenheit) at time t is given by [tex]T(t)=68+144e^{-0.04t}[/tex]

Initial Temperature for [tex]T(0)[/tex] can be calculated as

[tex]T(0)=68+144e^(0)=68+144=212[/tex]°[tex]F[/tex]

       [tex]=212[/tex]°[tex]F[/tex]

We can calculate, temperature after 15minutes as

[tex]T(15)=68+144e^(-0.04*15)=68+144e^(-0.6)=147[/tex]°[tex]F[/tex]

Time taken to reach soup at 125 degrees Fahrenheit can be calculated as:

[tex]125=68+144e^(-0.04t)[/tex]

⇒[tex]144e^(-0.04t)=125-68=57[/tex]

⇒[tex]t=23.3 minutes[/tex]

Hence, we can conclude that following is the correct answer.

Initial temperature of the soup is 212°F.

147°C is the the temperature of the soup after 15 minutes.

After 23.2 minutes soup will reach 125 degrees Fahrenheit.

Learn more about temperature here:

https://brainly.com/question/7510619?referrer=searchResults

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