# Video: Calculating the Final Temperature of an Object Cooling at a Continually Varying Rate

A bottle of water at 65°C is put in a refrigerator at 35°C. After an hour, the bottle has cooled to 50°C. What will the temperature of the bottle be after another hour?

05:24

### Video Transcript

A bottle of water at 65 degrees Celsius is put in a refrigerator at 35 degrees Celsius. After an hour, the bottle has cooled to 50 degrees Celsius. What will the temperature of the bottle be after another hour?

So we have this bottle of water at a certain initial temperature. And then we put the bottle of water inside a refrigerator at a different temperature. Now, if we close the door and then wait one hour, we find out the temperature of our bottle has decreased by 15 degrees to 50 degrees Celsius. Then the question is, if we close the refrigerator door once more and wait for another hour, then what will the temperature of the bottle be? Clearly, the bottle will keep cooling down until eventually it reaches the same temperature as the refrigerator.

Just how it does this is described mathematically by a relationship called Newton’s law of cooling. We can think about this law of cooling by realizing that it refers to objects that are at different temperatures. In general, there’s the temperature of the object we’re interested in, in our case the bottle. And secondly, there’s the temperature of the environment that the object is in, we can call it 𝑇 sub e. And in our case, that environment is the inside of the refrigerator.

This law of cooling tells us that if we take the difference between those two temperatures and then multiply that difference by an exponential factor raised to a constant value 𝑘 multiplied by the time 𝑡 that our object is in the environment and if we then add to this term the temperature of the environment, 𝑇 sub e, then, in that case, we have an equation for the temperature of our object as a function of the time passed, 𝑡.

This law of cooling then involves three physical parameters. It involves the temperature of our object, the temperature of the environment, and the time that our object spends in that environment.

Knowing all that, let’s apply Newton’s law of cooling to our particular scenario. As we said, in our application, the object we’re working with is this bottle of water which starts out at 65 degrees Celsius. We’ll call that 𝑇 sub o. And then, there’s our environment, the refrigerator, which is at a temperature of 35 degrees Celsius. We can name that 𝑇 sub e.

When we look back at this law of cooling to see what else in the equation we know, we notice in the exponent there’re these two factors 𝑘 and 𝑡. As we said, 𝑘 is a constant, which depends on the particular material we’re using. And 𝑡 is the time elapsed in seconds. For the first hour that the bottle cools, we’re told the resulting temperature. But we’re not told what that constant 𝑘 is.

Our Newton’s law of cooling equation then basically looks like this. 50 degrees Celsius, which is the temperature of the bottle after an hour has passed, is equal to the original temperature of the bottle minus the temperature of its environment times 𝑒 to the 𝑘 times 𝑡, where 𝑡 is one hour or 3600 seconds, plus the temperature of the environment, 35 degrees Celsius.

Notice that currently, we’re working in this Celsius temperature scale. In general, when we work with SI units, we would rather use the Kelvin scale because that’s the absolute temperature scale. Converting between Kelvin and degrees Celsius is a simple conversion process. But in this case, we don’t need to go through that process. Whether we work in Kelvin or degrees Celsius, our answer will be the same. That’s not always the case. But in this instance, it is. We’re okay then leaving this equation as is in degrees Celsius.

And we see that we want to solve for the constant 𝑘 in this exponent. To start doing that, let’s subtract 35 degrees Celsius from both sides of this equation, giving us on the left-hand side 50 minus 35 or 15 degrees Celsius. And then, we’ll divide both sides of the equation by 65 degrees Celsius minus 35 degrees Celsius. 65 minus 35 is equal to 30. And we see the left-hand side of our equation simplifies to one over two.

In order to solve for our value 𝑘, we’ll need to bring it down from the exponent, so to speak. And to do that, we can use a mathematical function called the natural logarithm. The natural logarithm has a property such that if we take the natural logarithm of 𝑒 to the 𝑥, then the result is the exponent 𝑥.

Therefore, if we take the natural logarithm of both sides of our equation, the natural log of 𝑒 to the 𝑘 times 3600 seconds, then on the right-hand side of our equation, we simply find 𝑘 times 3600 seconds. The constant 𝑘 then is equal to the natural log of one-half divided by 3600 seconds or negative 1.925 times 10 to the negative fourth inverse seconds. That’s the value for the constant 𝑘 for these objects.

Now that we know that constant, we’re ready to move on to the second hour, the one where the bottle starts at 50 degrees Celsius and cools for another 3600 seconds. This time, when we apply Newton’s law of cooling, our environment temperature, 𝑇 sub e, is once again 35 degrees Celsius. But now our object temperature, 𝑇 sub o, is 50 degrees Celsius. And we know the value of 𝑘. And we also know that we’re waiting another hour past the initial hour. It’s for that time, two hours past the initial time, that we want to know the temperature of our bottle of water.

Plugging in all these values, we find a result of 42.5 degrees Celsius. This is the temperature of our bottle of water after two hours of being in a 35 degrees Celsius refrigerator.