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Solve the given problems by integration.A hot metal rod with an initial temperature of $425^{\circ} \mathrm{C}$ is placed in a room with temperature $20^{\circ} \mathrm{C}$. The time $t$ (in min) required for the temperature $T$ of the rod to cool to $175^{\circ} \mathrm{C}$ is given by $$t=8.72 \int_{175}^{425} \frac{1}{T-20.0} d T$$.Find this time.

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 2

The Basic Logarithmic Form

Integrals

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in this Newton's law of cooling problem, We start off with the initial value and we're waiting to find out how long it will take to cool to a certain temperature. Now, a big part of this one is we're already given the K value, which is what we've had this all foreign previous problems. So that will save us some steps. But we'll start off with the original equation of the temperature as a function of time is going to equal be ah constant you and that is gonna cool but exponentially decreasing rate. So e to the negative k t bless the ambient temperature. And right away we could go ahead and pull again some initial values. So let's go ahead and pull again. 900 degrees is the temperature which will substitute on the left side. Vehicles Oh, are constant times the exponential raised to zero. Since we have a time of zero, it is the beat of the zero is the same thing as one and then plus R and B, a temperature of the room which is 30 degrees. This gives us our see that constant value, which we could see is gonna be 870 degrees civil Scroll down a little here, So that gives us see is 800 and 70. Next, we will use this in conjunction with the rest of the problem to set the whole thing equal to 100 would just use black this time. And we want to find out how long it will take before this rod reaches a temperature of 100. So we'll replace the left side with that people to again, we're gonna basically use this form of the equation. Um, so it is 8 70 times e. So the negative K, which is negative 0.2 times t which we do not know. So we could just go ahead and leave Battisti and then plus that ambient temperature of the room of 30 degrees tool sulfur t Here, um, we'll subtract the 30 over so little give us 70 and then from there we're gonna divide both sides by 8 70 or just combining a stuff or two. And that'll leave us in the end. On the right side, that's just e to the negative 0.0 to t now to solve for T, we would have to get rid of the exponential. So this is where we're gonna take the natural log of both sides. Now, one of the left equals a line of the right. And if we do take the natural log of of the exponential there in verses and so Bill cancel. And so we'll end up with Ln of 70 over 80 8 70 Divided by this is what's left over on the right side. So we'll divide both sides by mm. Negative. 0.2 The best by negative zero to. And I don't get finally just the time. And as we do, So, um, we're going to find out that this gives us about 126 minutes, so it will take about 126 minutes or a rod to cool down to that value of 100 degrees

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