Answer:
Step-by-step explanation:
Use Newton's Law of Cooling for this one. It involves natural logs and being able to solve equations that require natural logs. The formula is as follows:
[tex]T(t)=T_{1}+(T_{0}-T_{1})e^{kt}[/tex] where
T(t) is the temp at time t
T₁ is the enviornmental temp
T₀ is the initial temp
k is the cooling constant which is different for everything, and
t is the time (here, it's in minutes)
If we are looking first for the temp after 20 minutes, we have to solve for the k value. That's what we will do first, given the info that we have:
T(t) = 80
T₁ = 30
T₀ = 100
t = 5
k = ?
Filling in to solve for k:
[tex]80=30+(100-30)e^{5k}[/tex] which simplifies to
[tex]50=70e^{5k}[/tex] Divide both sides by 70 to get
[tex]\frac{50}{70}=e^{5k}[/tex] and take the natural log of both sides:
[tex]ln(\frac{5}{7})=ln(e^{5k})[/tex]
Since you're learning logs, I'm assuming that you know that a natural log and Euler's number, e, "undo" each other (just like taking the square root of something squared). That gives us:
[tex]-.3364722366=5k[/tex]
Divide both sides by 5 to get that
k = -.0672944473
Now that we have a value for k, we can sub that in to solve for T(20):
[tex]T(20)=30+(100-30)e^{-.0672944473(20)}[/tex] which simplifies to
[tex]T(20)=30+70e^{-1.345888946}[/tex]
On your calculator, raise e to that power and multiply that number by 70:
T(20)= 30 + 70(.260308205) and
T(20) = 30 + 18.22157435 so
T(20) = 48.2°
Now we can use that k value to find out when (time) the temp of the object cools to 35°:
T(t) = 35
T₁ = 30
T₀ = 100
k = -.0672944473
t = ?
[tex]35=30+100-30)e^{-.0672944473t}[/tex] which simplifies to
[tex]5=70e^{-.0672944473t}[/tex]
Now divide both sides by 70 and take the natural log of both sides:
[tex]ln(\frac{5}{70})=ln(e^{-.0672944473t})[/tex] which simplifies to
-2.63905733 = -.0672944473t
Divide to get
t = 39.2 minutes
