Answer :
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
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:
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