A pie is removed from an oven and its temperature is 175 ℃ 175℃ 175℃ and placed into a refrigerator whose temperature is constantly 3 ℃ 3℃ 3℃ . After 1 1 1 hour in the refrigerator, the pie is 90 ℃ 90℃ 90℃ . What is the temperature of the pie 4 4 4 hours after being placed in the refrigerator?

this problem, we're told that a pie is removed from an oven and its temperature is 175 degrees Celsius. It's then placed into a refrigerator whose temperature is constantly three degrees Celsius. After one hour, the pie has now cooled to 90 degrees. What is the temperature of the pie four hours after being placed in the refrigerator? So let's think about the information we're given here. We're given that initial temperature T of zero as 175 degrees Celsius. Then we're told that that refrigerator has a temperature that's constantly three. That's the temperature of our surrounding medium. Then we're told that t after one hour is equal to 90 degrees, and we're asked to find here t after four hours. So let's go ahead and get started with our steps here, starting by setting up our differential equation by plugging in that temperature of our surrounding medium. So we have a d big T over d little t is equal to k times t minus three. So here with that differential equation set up, we can go ahead and proceed here in separating our variables and integrating both sides. So I'm gonna divide both sides by t minus three here. That's going to cancel out. And at the same time, I'm gonna multiply by d, little t to get that to cancel. So now I have a dt over t minus three, just combining that into one fraction is equal to kdt. Now integrating both sides here on the left hand side, taking this integral gives me the natural log of the absolute value of t minus three, and that's then equal to kt plus c. So now that we've integrated both sides here, we need to find our constants. So we wanna figure out what k and c are. And as we proceed in step three, we're gonna start by finding c. So we need to plug in that initial condition that was given to us up here, that t of zero was equal to 175. So doing that here, we're plugging in zero for time and 175 for temperature. So this gives us here the natural log of 175 minus three is equal to k times zero plus C. This k times zero just goes to zero here and 175 minus three is 172. So we get here that the natural log of 172 is equal to c and we can plug that back into our equation here. So we can go ahead and replace that c. So this is now plus the natural log of 172. So with step three done, we proceed on to step four, which is now finding our other constant k by plugging in our other given temperature. Now our other given temperature was after one hour, and this was said to be 90 degrees. So t of one is equal to 90. And now we're plugging in one for time and 90 for temperature. So this is the natural log of 90 minus three and that's then equal to k times one plus the natural log of 172. So I just need to subtract this natural log of one seventy two over and then do this algebra ninety minus three is gonna be 87. So this is the natural log of 87 minus the natural log of one seventy two, and that's equal to k times one, which is just k. Now I can condense this down into one natural log, the natural log of 87 over one seventy two, and that gives me my constant k. Now again here, I've said this before, you can type this into a calculator and get a decimal, but keeping it in this form allows us to avoid rounding until we absolutely need to at the end. So I tend to keep it in this natural log form as long as possible. Now I can replace that k in my original equation with this. So moving that over, giving myself a little bit of space here, replacing that with the natural log of 87 over one seventy two. So now I have this equation with all of my constants known and I can go ahead and solve for t. So solving for t on this equation here, I'm just gonna rewrite my equation down here to begin with. So the natural log of t minus three is equal to the natural log of 87 over 172 times little t for time plus the natural log of 172. So now solving for t here, we wanna take e, raise it to each side here, that e and that natural log will cancel leaving us with t minus three. Then over here on this side, we can split this into two different natural logs. In doing that, we'll be able to cancel this one natural log and we'll end up with a 172 times e to the power of the natural log of 87 over 172 times little t time. Then last thing we need to do here is add three to both sides. That will allow this negative three to cancel over here and we now have our temperature as a function of time, but we're not quite done yet because we were asked to specifically find the temperature four hours after being in the refrigerator. So that means that we want to plug in four for that temperature or for that time T. So to find that value here we're looking for T of four. That's going to be equal to 172 times e to the power of the natural log of natural log of 87 over 172 times four, and then plus three. Note that this plus three is outside of that exponent. So now we want to type all of this into our calculator being really careful with parentheses so that we end up getting the right answer. And as we do this we'll end up getting here about 14.26 degrees Celsius for temperature after four hours. Now if you have any questions feel free to let us know in the comments, and I'll see you in the next one.

13: Intro to Differential Equations

Separable Differential Equations

Business Calculus

13: Intro to Differential Equations

Separable Differential Equations

Struggling with Business Calculus?

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Multiple Choice

A pie is removed from an oven and its temperature is $175℃$ and placed into a refrigerator whose temperature is constantly $3℃$ . After $1$ hour in the refrigerator, the pie is $90℃$ . What is the temperature of the pie $4$ hours after being placed in the refrigerator?

$226.1℃$

$214.7℃$

$14.26℃$

$13.69℃$

Verified step by step guidance

Step 1: Recognize that this problem involves Newton's Law of Cooling, which describes the rate of change of the temperature of an object as proportional to the difference between its temperature and the ambient temperature. The formula is typically written as:

T = T

_a+(T₀-T_a)e^(-kt), where

T

_a is the ambient temperature,

T

₀ is the initial temperature,

k

is the cooling constant, and

t

is time.

Step 2: Substitute the given values into the formula. The ambient temperature

T

_a is

3

, the initial temperature

T

₀ is

175

, and after

1

hour, the temperature

T

90

. Plug these values into the formula to solve for the cooling constant

k

90 = 3 + (175 - 3) e^(- k)

Step 3: Solve for

k

. Rearrange the equation to isolate the exponential term:

(90 - 3) / (175 - 3) = e^(- k)

. Take the natural logarithm of both sides to solve for

k

k = - \ln ((90 - 3)/ (175 - 3))

Step 4: Use the value of

k

to find the temperature of the pie after

4

hours. Substitute

k

t = 4

, and the other known values into the formula:

T = 3 + (175 - 3) e^(- k (4))

Step 5: Simplify the equation to find the temperature of the pie after

4

hours. This involves calculating the exponential term and adding it to the ambient temperature. The result will be the temperature of the pie at that time.