Question

An object is heated to 100°. It is left to cool in a room that

Answer 1

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:

$T(t)=T_{1}+(T_{0}-T_{1})e^{kt}$ 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:

$80=30+(100-30)e^{5k}$ which simplifies to

$50=70e^{5k}$ Divide both sides by 70 to get

$\frac{50}{70}=e^{5k}$ and take the natural log of both sides:

$ln(\frac{5}{7})=ln(e^{5k})$

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:

$-.3364722366=5k$

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):

$T(20)=30+(100-30)e^{-.0672944473(20)}$ which simplifies to

$T(20)=30+70e^{-1.345888946}$

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 = ?

$35=30+100-30)e^{-.0672944473t}$ which simplifies to

$5=70e^{-.0672944473t}$

Now divide both sides by 70 and take the natural log of both sides:

$ln(\frac{5}{70})=ln(e^{-.0672944473t})$ which simplifies to

-2.63905733 = -.0672944473t

Divide to get

t = 39.2 minutes