Newton's Law ef Cooling Boiling water, at 100^∘ C, is placed in a freezer at 0^∘ C. The temperan

Newton's Law ef Cooling Boiling water, at 100^∘ C, is placed...

Newton's Law ef Cooling Boiling water, at $100^{\circ} \mathrm{C}$, is placed in a freezer at $0^{\circ} \mathrm{C}$. The temperanure of the water is $50^{\circ} \mathrm{C}$ after 24 min. Find the temperature of the water after
96 min. (Hint: Change minutes to hours.)

Key Concepts

Newton's Law of Cooling

This concept states that the rate at which an object's temperature changes is proportional to the difference between its current temperature and the ambient temperature. It serves as the basis for modeling the cooling or heating of objects in environments with constant surrounding temperatures.

Exponential Decay

In the context of Newton's Law of Cooling, the difference between the object's temperature and the ambient temperature decreases exponentially over time. This decay is represented mathematically by an exponential function that involves a decay constant, illustrating how quickly the temperature approaches the ambient temperature.

Initial and Ambient Conditions

Accurate application of Newton’s Law requires identifying the initial temperature of the object and the constant ambient temperature. These values serve as boundary conditions that are essential for solving the differential equation describing the cooling process.

Time Conversion and Consistency

Proper application of the cooling model necessitates consistent units of time, ensuring that any time measurements (such as minutes or hours) align with the rate constant and the form of the exponential decay equation. This step is critical for accurate computation of temperature changes over time.

Transcript

00:01 We're using newton's law of cooling, and we know that the temperature as a function of time will be the temperature of the medium or that initial temperature that you're putting the object in.

00:13 And we know that the temperature is zero degrees celsius.

00:20 So this initial temperature of the freezer is zero degrees celsius.

00:25 And we know that at time and what the temperature will be, or f of t, that at time zero, the object is put in there and it's 100 degrees celsius.

00:41 And then after 24 minutes, so of our time we're in minutes, we would have 24.

00:48 Now, they suggest that you change it to hours, and if we take 24 divided by 60, you get 0 .4 hours.

00:54 It actually doesn't matter which you do, but that after 0 .4 hours or 24 minutes, the temperature has dropped to 50 degrees celsius.

01:04 And we want to know after 96 minutes, which is 1 .6 hours, what the temperature will be.

01:13 So that's what we want to find.

01:15 So we know that f of 0 is equal to 100 degrees.

01:24 So we know that that's what the temperature is.

01:27 And we know that if we plug our time of zero in here, that this will become zero, and e to the 0th will become 1.

01:39 And so we'll find that the c value is equal to 100 degrees celsius.

01:44 So so far we know the function is this.

01:47 F of t is equal to the 0, which we don't even have to write down, plus the 100 times 8.

01:55 To the negative k times t so we've used that piece of information now we'll go to this piece of information this will allow us to find the k and so let's substitute in the 0 .4 hours we'll do what they suggest and do our time in terms of hours and we have a hundred times e to the 0 .4 times negative k and that that needs to equal 50 degrees celsius and so dividing in.

02:27 So remember our time we're going to be doing this because of what i did here in terms of hours.

02:33 So when we divide both sides by 100, we'll end up getting 0 .4k is equal to 0 .5...

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