Newton's Law of Cooling. According to Newton's law of...

Newton's Law of Cooling. According to Newton's law of cooling, if an object at temperature $ T $ is immersed in a medium having the constant temperature $ M $, then the rate of change of $ T $ is proportional to the difference of temperature $ M-T $. This gives the differential equation $$ d T / d t=k(M-T) $$ (a) Solve the differential equation for $ T $ (b) A thermometer reading $ 100^{\circ} \mathrm{F} $ is placed in a medium having a constant temperature of $ 70^{\circ} \mathrm{F} $. After 6 min, the thermometer reads $ 80^{\circ} \mathrm{F} $. What is the reading after 20 min? (Further applications of Newton's law of cooling appear in Section 3.3.)

Newton's Law of Cooling. According to Newton's law of cooling, if an object at temperature $ T $ is immersed in a medium having the constant temperature $ M $, then the rate of change of $ T $ is proportional to the difference of temperature $ M-T $. This gives the differential equation
$$ d T / d t=k(M-T) $$
(a) Solve the differential equation for $ T $
(b) A thermometer reading $ 100^{\circ} \mathrm{F} $ is placed in a medium having a constant temperature of $ 70^{\circ} \mathrm{F} $. After 6 min, the thermometer reads $ 80^{\circ} \mathrm{F} $. What is the reading after 20 min? (Further applications of Newton's law of cooling appear in Section 3.3.)

Key Concepts

Exponential Decay in Differential Equations

Exponential decay is a concept where a quantity decreases at a rate proportional to its current value, resulting in a solution that is an exponential function. In problems involving Newton's Law of Cooling, the solution is an exponential function showing how the temperature difference between the object and the medium decays over time.

Separation of Variables

Separation of variables is a method for solving differential equations by rearranging the equation so that each variable appears on a different side of the equation. This technique is particularly useful for solving equations like those arising from Newton’s Law of Cooling, allowing integration of both sides to find a solution.

Newton's Law of Cooling

Newton's Law of Cooling describes the process by which an object exchanges heat with its surroundings. In particular, the law asserts that the rate of temperature change of an object is proportional to the difference between its own temperature and that of the surrounding medium. This principle is used to model how objects cool or warm towards the ambient temperature.

First-Order Linear Differential Equations

A first-order linear differential equation involves a function and its first derivative, and can be written in a standard form where the derivative is proportional to the function plus a function of the independent variable. In the context of Newton’s Law of Cooling, the equation is linear and its solution typically involves exponential functions.

Transcript

00:01 So we're given the differential equation of d t over t equal to k of m minus t.

00:10 I write, we write this as d t over d t equal to minus k.

00:18 I pull out a minus and i have t minus m.

00:22 This is just the same thing.

00:24 Now i separate these and i have d t over t minus.

00:30 Minus m equal to minus k multiplied by d t now let's take the integral of both sides you will have that the integral of this will just be the natural log of absolute value of t minus m equal to when you integrate these k is a constant so you have minus k t and then let me add plus c one the next thing we do is that we take the exponential of both sides.

01:01 So i have e and i have e.

01:04 So that gives me that this will cancel out.

01:07 I have that t minus m equal to e to the minus kt multiply by e to the c1, which is equal to if i take this e to the c1 is a constant, so i can name that c2.

01:23 So i have c2, e to the minus kt.

01:27 So that implies that absolute value of t minus m equal to c2 e minus k t, which implies that t minus m equal to plus or minus of c2 e minus kt.

01:44 This implies that t equal to m plus c e minus k t, where c equal to plus c e minus k t, where c equal to plus a minus of.

01:57 Of c2.

01:59 So let's name this equation star.

02:01 We're going to use it later on.

02:03 Now the b parts we're given that the constant temperature m equal to 70 degrees.

02:12 We're given t at zero equal to 100 degrees.

02:17 We're given t at 6 equal to 80 degrees and we want to find t at 20.

02:26 That's what we're looking for.

02:28 Now plug this into equation star, plugging into star, the equation star we got.

02:36 So we have that t at zero, that's 100 equal to 70.

02:43 Our m is 70 plus c, e to the, because at t equals to zero, that's e to the zero...

Please give Ace some feedback