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The answers to most of the following exercises are in terms of logarithms and exponentials. A calculator can be helpful, enabling you to express the answers in decimal form.The temperature of an ingot of silver is $60^{\circ} \mathrm{C}$ above room temperature right now. Twenty minutes ago, it was $70^{\circ} \mathrm{C}$ above room temperature. How far above room temperature will the silver bea. 15 min from now?b. 2 hours from now?c. When will the silver be $10^{\circ} \mathrm{C}$ above room temperature?

a.) $53^oC$ b.)$24^o C$c.) 233 min

Calculus 1 / AB

Calculus 2 / BC

Chapter 7

Integrals and Transcendental Functions

Section 2

Exponential Change and Separable Differential Equations

Functions

Trig Integrals

Missouri State University

Harvey Mudd College

Baylor University

University of Michigan - Ann Arbor

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So we have this sea over India. That right now is both the room temperature by 60 degrees. And 20 minutes ago, it was 70 degrees above. What we want to do is determine how far above the room temperature this silver in it will be in 15 minutes. And to our shore mount as well as win, Wouldn't silver ring it be 10 degrees above our room temperature? All right, so let's first create two equations that we can use. So this first equation we can see if we can get something from here. So it says our time is zero, and it says 60 degrees above room temp. So this is a way of saying that it's a TSH minus our surrounding temperature because we don't exactly know what they're room temperature is. Um, but all we care about is that's the difference, which is going to be that left hand side So you could have this equation here, so it's 60 is equal to Well, we don't know what the initial temperature is now. We don't know our surrounding, and then we will end up with when we plug in T E to zero, which is just going to be one. So we actually have. That's the difference in our initial and surrounding temperature should be 60. All right, so that's homes pretty good so far, because that means we can just replace this with 60 cause that's just a constant. Now let's see what else we can get when we plug stuff into this second equation. So if we plug in to e substitution that we made up here already, so 70 of Bo is going to be 70 on the left hand side. We already said that it's not minus hs should be 60 and then we have e to the negative K. And since it's had 20 minutes ago, I'm gonna use negative 20 minutes for this. So now the negatives will counts out with each other, and we just need to solve for Kay. So we would divide each side by sixties, were gonna get 7/6, and we would want to take the natural log on each side. That's gonna give us 20 times K. Now to get kay, we're gonna divide each side by 20 so have won over 20 times the natural log of 76 and then uses about a little, and the units with fish should be reciprocal minutes. And that will be important, especially since a heart be there talking about ours as opposed to minutes. But let's see what this gives us approximately. So natural. Log of seven, divided by six, is about one point. I multiply that by 1/20 so that should give something around zero point 0077 Reciprocal minutes aren't so. That's Kay. Now we can start solving or our difference on each side. So for part, pay all we're going to need to do rectory. Let me move all of this down here and I can zoom back in again. That way we'll be a little bit easier to actually see what I'm doing. Looks block all that off. So for part A So we want to see what h minus hs is when tea is 15. Well, this is going to be 60 times he too negative 0.77 and then we plug in 50 which looks like it will give something around. So we would want to exponentially eight those which is about 0.89 and we multiply that by 60 so it looks like the difference will be around 53 degrees Celsius. So that will be our answer for part. Hey, and now, for part be well, since our rate constant is in reciprocal minutes, we want to make sure that our time unit is also in minutes. So we were just multiply that by 1 20 to get 1 20 minutes. So then h minus are surrounding Should be 60 times he to the negative 0.77 times 120. And this should give us. So before it was 0.8 times zero dates. When we exponentially ated that actually that was with other number we need to multiply that by 1 20 this time. So 1 20 times negative zero 077 sets about negative 9.24 Now we exponentially ate that number, which gives us about 0.39 Multiply that by 60. So that's going to give us something. Let me write this story So this is going to be approximately 24 degrees Celsius, so that would be the difference in the temperature after two hours. Now, for part C, we want to set the left hand side equal to 10 and then soul for time. So we're gonna divide each side by 60 circuit, but it puts. See, appear so 1/60. Take the natural log and we'll get negative 0.7 17. So T is going to be equal to, and I'll rewrite natural ago. One over six has negative natural log of six. And then we're gonna divide over to get t bite. So? So the negatives counts out with each other, and so this is going to be approximately equal to the natural log of six, divided by 0.77 which is going to be 230 two are about 233. And the units with this should be in minutes since our rate constant was in reciprocal minutes. So that's about how long it would take for the difference to be 10 degrees

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