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Answer :
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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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
