A container of hot liquid is placed in a freezer that is kept at a constant temperature of 20^∘

A container of hot liquid is placed in a freezer that is...

Key Concepts

Newton's Law of Cooling

Newton's Law of Cooling states that the rate at which an object cools (or heats) is proportional to the difference between the object's temperature and the ambient temperature. This principle forms the basis for modeling temperature changes over time with an exponential function.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. In cooling problems, the difference between the object's temperature and the surrounding temperature diminishes exponentially over time.

Differential Equations

Differential equations provide a mathematical framework for modeling processes that change continuously, such as temperature variations in cooling. Newton’s Law of Cooling is typically expressed as a first-order differential equation whose solution reveals the exponential relationship between temperature and time.

Logarithmic Functions

Logarithmic functions are crucial for solving exponential equations derived from differential equations. They are used to isolate the time variable when determining how long it will take for a temperature to reach a certain value, as the exponent in the exponential decay model must be solved using logarithms.

Transcript

00:01 In this question, we are given that a hot liquid is put into a freezer or refrigerator, and the freezer temperature is 20 degrees fahrenheit.

00:12 At t equals to zero, the temperature of the liquid is 160 degrees fahrenheit.

00:18 When t equals to five minutes, that means five minutes later, the temperature is 60 degrees fahrenheit.

00:24 We want to find how much longer will the temperature of this liquid become 30 degrees fahrenheit.

00:30 High.

00:31 Now before we do this, let's look at some basics you require.

00:35 If i have a dy over dt is equals to a constant multiplied to a function of y.

00:47 To solve this differential equation, i need to bring the function of y over and leave the constant on the right side.

00:57 Then apply an integral on both sides with respect to t.

01:02 So on the left, you realize that you are just integrating one over function of y respect to y and on the right side you're just integrating k a constant respect to t the integration rules you require will be like power rule if i have t to power of n and i integrate that respect to t i get t to power of n plus 1 over n plus c a special case will be just integrating k constant now i can put t to power of 0 here like this so using the power rule, i'll just get kt plus c.

01:41 Now, if i were to integrate one over ay plus b, the a and b are just constants, with respect to y, i'll get lone mod of a plus b divided by the a plus c.

02:00 Okay, so now let's start.

02:02 We let y be the temperature at time t, temperature of the liquid at time t.

02:15 Then newton's law of cooling says that, says that the d .y, dx or y prime is equal to a constant k, multiplied to the y, the difference in the temperature of the environment.

02:38 So it's y minus 20 degrees fahrenheit.

02:43 So now, let's bring this guy to the left side.

02:50 And the y prime, i can write it as dy over dt.

02:56 It's equals to k on the right side.

03:00 Apply an integral on both sides with respect to t.

03:07 You can see on the left, i'm integrating 1 over 1 minus 20 respect to y.

03:12 So i'll just get lawn mod of y minus 20.

03:17 On the right, integrating k respect to t.

03:20 I'll just get kt plus let's say a, where a is a constant.

03:26 Now let's take e on both sides.

03:35 Now, on the right side, i can split it into e to the power of kt, times e to power a.

03:42 That's using indices.

03:44 And notice that no matter what p and a is, as long as you take exponential to the power of these values, is always gradient zero.

03:57 That means on my left side, on my left side, i'm just going to have mod of 1 minus 20, right? because e to the long of that is just going to be more of y minus 20.

04:09 Now, since the whole expression is great and 0, i can drop the mod...

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