43. Surrounding medium of unknown temperature A pan of warm wa...

Question

43. Surrounding medium of unknown temperature A pan of warm water (46°C) was put in a refrigerator. Ten minutes later, the water’s temperature was 39°C; 10 min after that, it was 33°C. Use Newton’s Law of Cooling to estimate how cold the refrigerator was.

Accepted Answer

Hello. In this video, we are told that a thermometer reads 25 °C before being placed in an oven. After 4 minutes, the temperature reads 40 °C, and after 8 minutes, it reads 52 °C. We want to use Newton's law of cooling to estimate the oven's temperature. So in order to approach this problem, let's go ahead and review Newton's law of cooling. Newton's law of cooling equation is a function T in terms of temperature. That is equal to the temperature of the surrounding area TS plus the quantity of our initial temperature minus the temperature of the surrounding area multiplied by E raised to the power of KT. Here, k represents the cooling or heating coefficient, and t represents time. So, Let's go ahead and read the given information that the problem provides us. The first part of the problem says that the temperature reached 25 °C before being placed in an oven. What this means is that our initial temperature t_zero is going to equal to 25 °C. Now, this part here tells us that after 4 minutes, the temperature is reading at 40 °C. So what this means is that our function T at 4 minutes is going to equal to 40 °C. And then we are told that after 8 minutes, the temperature reads at 52 °C. So the output of our temperature function at 8 minutes is going to equal to 52 °C. So what we can go ahead and do is we can set up a system of equations by using the given information. So the first thing we're going to do is we're going to plug in our initial value of 25 °C into our temperature function. That is going to give us a general temperature function T T equal to TS plus 25 minus TS multiplied by E raised to the power of KT. Now here, if we plug in our first condition of T4 into this equation, we can come up with the equation of 40 because that's the output of the function, 40 equal to TS plus the quantity 25 minus TS multiplied by e raises the power of -4T. And here, T gets replaced with -4 because that is the value of T for this initial condition. Repeating this process for the second given condition gives us the second equation 52 degrees equal to ts plus 25 minus ts multiplied by e raises the power of -8 T. So this is going to be our system of equations, but let's go ahead and try to find a way to simplify our exponential forms. Let's go ahead and create a substitution for Ease the power of -4T. We can let x equal to e raise the power of -4T. But notice that E raised the power of -4T can also be squared, so we have E raised the power of -4T, quantity squared, and that is also equivalent to x 2. So what this means is that if we use X is equal toe raised the power of -4T, this first exponential statement in our first equation replaces with X, and equivalently, the second exponential value will be replaced with X 2. So we can further rewrite our system as 40 equal to TS plus the quantity 25 minus TS multiplied by X. And the 2nd equation will be 52. Equal to ts plus 25 minus ts multiplied by x 2. Now, what we're going to do in this case is we're going to solve for X in the first equation. If we solve for x, we get x is equal to 40 minus ts divided by 25 minus ts. And if we were to take this value of x and plug it into our second equation by substituting, we can rewrite the second equation as 52 equal to TS plus the quantity 25 minus TS multiplied by 40 minus TS divided by 25 minus ts quantity squared. We can now simplify this equation to try to solve forts. So we will rewrite. The quantity 40 minus ts divided by 25 minus ts square by using our exponential properties. We will distribute the exponent to both the numerator and denominator, and that will allow us to rewrite this quantity as 40 minus ts quantity squared divided by 25 minus ts quantity squared. Now, if we were to multiply this, but with the 2nd term in the equation, we can eliminate one of the factors from the numerator and denominator. So, our equation is going to simplify to be 52 equal to TS plus 40 minus TS quantity squared divided by 25 minus TS. Now here what we're going to do is we're going to subtract TS to the left hand side and multiply both the equation, both sides of the equation by the denominator. That will leave us with 52 minus TS multiplied by 25 minus TS and this will equal to 40 minus TS quantity squared. Now, in order to further simplify this equation, we are going to expand both sides of the equation. So, simplifying the left-hand side will give us 1300. -52 TS minus 25 TS plus TS 2. And this is going to equal to 1600 minus 80 TS plus TS2. Now here, because we have the same square term on both sides of the equation, we can eliminate them by subtracting the values. Then we're going to combine like terms on the left-hand side of the equation. So we can further rewrite this equation to be 1300. 77 ts and this is going to equal to 1600. -80 ts What we can go ahead and do is move all the TS terms to the left hand side of the equation and all the constants to the right. So, we will have 80 TS minus 77 TS equal to 1600 minus 1300. Combining like terms will simplify this to be 3 multiplied by ts equal to 300. And all we have to do now is just divide by 3, and we will get an estimated temperature of 100 °C. This is going to be the solution to the given problem. And the overall solution is going to be option D. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

43. Surrounding medium of unknown temperature A pan of warm water (46°C) was put in a refrigerator. Ten minutes later, the water’s temperature was 39°C; 10 min after that, it was 33°C. Use Newton’s Law of Cooling to estimate how cold the refrigerator was.

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