A loaf of bread is removed from an oven at 350 degrees F and cooled in a room whose temperature is 70 degrees F. If the bread cools to 210 degrees F in 20 minutes, how much longer will it take the bread to cool to 155 degrees F. Round to the nearest minute.

As you know, Newton's law says that your Temperature function is

T(t) = 70 + 280e^-kt

So, just solve for k in

70 + 280e^-20k = 210

then solve for t in using T(t) = 155.

To find out how much longer it will take the bread to cool to 155 degrees F, we need to determine the cooling rate of the bread.

The bread cools from 350 degrees F to 210 degrees F in 20 minutes, which is a temperature change of 350 - 210 = 140 degrees F.

Let's calculate the cooling rate per minute:
Cooling rate = temperature change / time = 140 degrees F / 20 minutes = 7 degrees F per minute.

Now, we need to determine the remaining temperature change required to cool the bread to 155 degrees F:
Remaining temperature change = 210 degrees F - 155 degrees F = 55 degrees F.

Finally, we can calculate how much longer it will take to cool the bread to 155 degrees F:
Time required = remaining temperature change / cooling rate = 55 degrees F / 7 degrees F per minute ≈ 7.86 minutes.

Rounding to the nearest minute, it will take approximately 8 minutes longer for the bread to cool to 155 degrees F.

To find out how much longer it will take for the bread to cool to 155 degrees F, we can use Newton's Law of Cooling.

Newton's Law of Cooling states that the rate at which an object cools is proportional to the temperature difference between the object and its surroundings.

In this case, we have:

Initial temperature of the bread (T0) = 350 degrees F
Temperature of the room (Troom) = 70 degrees F
Temperature of the bread after 20 minutes (T20) = 210 degrees F
Temperature we want to find (Tfinal) = 155 degrees F

Using the formula for Newton's Law of Cooling:

T(t) = Troom + (T0 - Troom) * e^(-kt)

where T(t) is the temperature at time t, Troom is the temperature of the room, T0 is the initial temperature, k is the cooling constant, and e is the base of the natural logarithm.

We can rearrange the formula to solve for the cooling constant:

k = ln((T0 - Troom) / (T(t) - Troom)) / t

Using the information given, we can calculate the cooling constant for the bread:

k = ln((350 - 70) / (210 - 70)) / 20 = ln(280 / 140) / 20 = ln(2) / 20

Now, we can substitute the values into the formula to find out how much longer it will take for the bread to cool to 155 degrees F:

155 = 70 + (350 - 70) * e^(-((ln(2) / 20) * t))

Simplifying the equation:

85 = 280 * e^(-ln(2) * t / 20)

Dividing both sides by 280:

0.303571429 = e^(-ln(2) * t / 20)

Taking the natural logarithm of both sides (ln(e^x) = x):

ln(0.303571429) = -ln(2) * t / 20

Solving for t:

t = -20 * ln(0.303571429) / ln(2)

Using a calculator, we find:

t ≈ 33.84 minutes

Therefore, it will take approximately 34 minutes (rounding to the nearest minute) for the bread to cool from 210 degrees F to 155 degrees F.