27–30. Newton’s Law of Cooling Solve the differential equation...

Question

27–30. Newton’s Law of Cooling Solve the differential equation for Newton’s Law of Cooling to find the temperature function in the following cases. Then answer any additional questions.

An iron rod is removed from a blacksmith’s forge at a temperature of 900°C . Assume k=0.02 and the rod cools in a room with a temperature of 30°C When does the temperature of the rod reach 100°C?

Accepted Answer

Welcome back, everyone. In this problem, a cup of tea is initially at a temperature of 85 °C and cools in a room maintained at 20 °C. If the cooling constant is K equals 0.03 per minute, after how many minutes will the tea reach 40 °C? Run their answer to the nearest whole number. A says it's 25 minutes, B 32 minutes. C 39 minutes, and D 51 minutes. Now what do we know that will help us to figure out after how many minutes the tea will reach 40 °C? Well, here we're talking about cooling, right, and recall that by Newton's law of cooling or change in temperature T with respect to time T equals negative K or cooling constant multiplied by our temperature T minus the temperature of the room. So in this case, if we apply what we know here that uh our room is maintained at 20 °C and K equals 0.03 per minute. Then that means DT with respect to T is going to be equal to -0.03 multiplied by T minus 20. Which means then that 1 divided by t minus 20. DT equals -0.03 multiplied by DT or change in time. And know if we're going to solve for our temperature, OK, then that means we can integrate both sides. So we can integrate with respect to our temperature on the left. Let me put that in red here to show what we're doing. And then we can integrate with respect to time on our right. OK. Let me just make some space here to make it a little bit more clear. Now, if we think about it, when we integrate, OK, let me come over to the left here. On the left hand side, we're going to have the natural log of T minus 20, OK. And on the right hand side we're going to have negative 0.03 T plus C. So now we could use our initial condition to solve for our constant C and then we can plug in our constant to help us solve for our time T. So now in this case, OK. At, remember we're told again in our problem that T is initially at a temperature of 85 °C. OK. So at a time of 0, the temperature equals 85 degrees. OK. So this implies that T of 0, OK, equals 85. I know if we put that into our problem. OK, if we put that into our expression, this means that the natural log of 85 minus 20. Is going to be equal to. Uh, -0.03 multiplied by 0, which would be 0, plus C. 85 minus 20 is 65, so this means that C equals LN 65. Now that we have that, we can solve for our time T when our temperature T equals 40. So now, let me put this in red. When T is equal to 40, plugging this back into our equation, then this must mean. The the natural log of 40 minus 20 using our equation that we had here earlier, OK. Well, I didn't want an arrow actually. Let me just highlight it here. So the natural log of 40 minus 20 will be equal to -0.03 T. Plus the natural log of 65 LN 65. Now 40 minus 20 equals 20, so this means then. That negative 0.03 T equals LN20. Minus 165. And now in that case, This means then that. Here negative 0.03 T equals LN 20 divided by 65 or LN4 LN of 4/13, OK. And if negative 0.03 T equals LN of 4/13, then that means T would be equal to LN of 4/13 divided by -0.03. And that, OK, when we work it out would be approximately equal to 39 to the nearest minute. Thus, the time it takes for the tea to reach 40 °C or to cool to 40 °C will be approximately 39 minutes. If we look back at our answer choices, that means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Newton’s Law of Cooling

Solving First-Order Differential Equations

Exponential Decay and Time to Reach a Specific Temperature

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