Animated Solution for Physics - Properties of Solids and Liquids: In $5\text{ min}$, a body cools from $75^{\circ }\text{C}$ to $65^{\circ }\text{C}$ at room temperature of $25^{\circ }\text{C}$. The temperature of body at the end of next $5\text{ min}$ is ......... ${}^{\circ }\text{C}$.

Have you ever noticed how a piping hot cup of coffee cools down rapidly at first, but then seems to take forever to reach room temperature? This isn't just an illusion; it's a fundamental law of nature.

Sir Isaac Newton, the same genius who gave us the laws of motion, also formulated the Law of Cooling. He discovered that the rate at which an object loses heat is directly proportional to the temperature difference between the object and its surroundings.

In this problem, we are going to use a highly practical approximation of this law to predict the future temperature of a cooling body.

When dealing with small temperature changes over short time intervals, we can use the average temperature form of Newton's Law of Cooling.

The equation is beautifully simple:

Here, the left side represents the average rate of cooling. The right side contains our cooling constant $k$, multiplied by the difference between the average temperature of the body during that interval and the constant room temperature $T_s$.

Our first mission is to find the unique cooling constant $k$ for this specific body. We are given that in the first $5$ minutes, the temperature drops from $75^{\circ }\text{C}$ to $65^{\circ }\text{C}$ in a room that is $25^{\circ }\text{C}$.

Let's substitute these values into our master equation:

Simplifying the left side, the rate of cooling is $2^{\circ }\text{C}$ per minute. On the right side, the average temperature is $70^{\circ }\text{C}$. Subtracting the room temperature gives us $45^{\circ }\text{C}$.

Solving for $k$, we get:

This constant is the "DNA" of our cooling body. It dictates exactly how it will behave in the future.

Now comes the exciting part. We need to find the temperature of the body at the end of the next $5$ minutes.

The body starts this new interval at $65^{\circ }\text{C}$ and will cool down to an unknown final temperature, which we will call $T_3$. The time interval is again $5$ minutes, and the room temperature remains $25^{\circ }\text{C}$.

We set up our equation once more, this time armed with our cooling constant:

This equation might look a bit intimidating, but let's break it down step by step. First, we can simplify the fractions by canceling the $5$ on the left with the $45$ on the right, leaving a $9$ in the denominator.

Notice how the $2$ in the numerator cancels perfectly with the $2$ in the denominator of the average temperature term. This leaves us with:

To clear the fraction, we cross-multiply by $9$:

Expanding the left side gives:

Now, it's just a matter of simple algebra. We bring all the $T_3$ terms to one side and the constants to the other:

Dividing by $10$, we arrive at our final answer:

Notice something fascinating here. In the first $5$ minutes, the temperature dropped by $10^{\circ }\text{C}$. In the next $5$ minutes, it only dropped by $8^{\circ }\text{C}$. This perfectly aligns with Newton's Law: as the body gets closer to room temperature, it cools down more slowly!