High School

A pot of boiling soup with an internal temperature of 100 degrees F was taken off the stove to cool in a 69 degrees F room. After 15 minutes, the internal temperature of the soup was 95 degrees F. To the nearest minute, how long will it take the soup to cool to 80 degrees F?

Answer :

Answer:

86.34 min

Step-by-step explanation:

Using the formula for Newton's law of cooling.

[tex]T = T_{1} + (T_{0} - T_{1})e^{kt}[/tex]

where T₀ = initial temperature of object = 100 F

T₁ = temperature of surrounding = 69 F

T = final temperature of object and

t = time

Initially, when the temperature of the soup drops to 95 F, it takes 15 minutes. So, T = 95 F and t = 15 min

Substituting all the variables into the equation for T we have

[tex]95 = 69 + (100 - 69)e^{15k}\\95 - 69 = 31e^{15k}\\26 = 31e^{15k}\\e^{15k} = 26/31\\e^{15k} = 0.8387\\15k = ln(0.8387)\\k = ln(0.8387)/15\\k = -0.1759/15\\k = -0.012[/tex]

So,

[tex]T = 69 + (100 - 69)e^{-0.012t}\\T = 69 + 31e^{-0.012t}[/tex]

When T = 80 F,

[tex]80 = 69 + 31e^{-0.012t}\\ 80 - 69 = 31e^{-0.012t}\\11 = 31e^{-0.012t}\\e^{-0.012t} = 11/31\\e^{-0.012t} = 0.3548\\-0.012t = ln(0.3548)\\t = -ln(0.3548)/0.012\\t = 1.0361/0.012\\t = 86.34[/tex]

So, it takes the soup 86.34 minutes to cool to 80 °F

It will take 45 minutes for the soup to cool to 80 ° F.

Since a pot of boiling soup with an internal temperature of 100 degree F was taken off the solve to cool in a 69 degree F room, and after 15 minutes the internal temperature of the soup was 95 degree F, to determine how long will it take the soup to cool to 80 degree F the following calculation should be performed:


  • (100 - 95) / 15 = Degrees decreased per minute
  • 5/15 = X
  • 0.333 = X


  • (95 - 80) / 0.333 = X
  • 15 / 0.333 = X
  • 45 = X


Therefore, it will take 45 minutes for the soup to cool to 80 ° F.

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