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.)
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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.