An object is heated to 100°. it is left to cool in a room that
has a temperature of 30°. after 5 minutes, the temperature
of the object is 80°. what is the temperature of the object
after 20 minutes? when will the temperature of the object
be 35°?

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:

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:

which simplifies to

Divide both sides by 70 to get

and take the natural log of both sides:

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:

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):

which simplifies to

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 = ?

which simplifies to

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

which simplifies to

-2.63905733 = -.0672944473t

Divide to get

t = 39.2 minutes

answer: 270

step-by-step explanation:

you have to times 200 by 1.35 to get your answer

what is the your question?