A container of hot liquid is placed in a freezer that is kept at a constant temperature of 10°F. The initial temperature of the liquid is 160°F. After 5 minutes, the liquid's temperature is 60°F. How much longer will it take for its temperature to decrease to 20°F? (Round your answer to two decimal places.)

Have you learned about Newton's Law of Cooling?

To solve this problem, we can use the concept of exponential decay. The general formula for exponential decay is:

y = a * e^(-kt)

Where:
y is the final amount or temperature,
a is the initial amount or temperature,
e is Euler's number (approximately 2.71828),
k is the decay constant,
and t is the time elapsed.

In this case, we want to find out how much longer it will take for the liquid's temperature to decrease from 60°F to 20°F. Let's call this unknown time "x".

Using the given information, we can set up the following equation:

20 = 60 * e^(-kx)

To solve for the decay constant k, we can rearrange the equation:

e^(-kx) = 20 / 60
e^(-kx) = 1/3

Now, take the natural logarithm (ln) of both sides to eliminate the exponential:

ln(e^(-kx)) = ln(1/3)
-kx * ln(e) = ln(1/3)
-kx = ln(1/3)
kx = -ln(1/3)

Next, we need to find the value of k. To do this, we can use the initial temperature and time given:

60 = 160 * e^(-5k)

We can rearrange this equation to solve for k:

e^(-5k) = 60 / 160
e^(-5k) = 3/8

Take the natural logarithm (ln) of both sides:

ln(e^(-5k)) = ln(3/8)
-5k * ln(e) = ln(3/8)
-5k = ln(3/8)
k = -ln(3/8) / 5

Now that we have the value of k, we can substitute it back into the equation we got earlier:

kx = -ln(1/3)
(-ln(3/8) / 5) * x = -ln(1/3)

We can solve for x by isolating it:

x = (-ln(1/3)) / (-ln(3/8) / 5)

Simplifying further:

x = (-ln(1/3)) * (5 / -ln(3/8))
x = ln(1/3) * (5 / ln(3/8))

Now, we can calculate the value of x:

x ≈ ln(1/3) * (5 / ln(3/8))
x ≈ -0.847 * (5 / -0.693)
x ≈ -0.847 * -7.225

Rounding the answer to two decimal places:

x ≈ 6.13 minutes

Therefore, it will take approximately 6.13 minutes for the liquid's temperature to decrease to 20°F.