Animated Solution for Physics - Properties of Solids and Liquids: A body takes $4\text{ min}$ to cool from $61^{\circ }\text{C}$ to $59^{\circ }\text{C}$. If the temperature of the surroundings is $30^{\circ }\text{C}$, then the time taken by the body to cool from $51^{\circ }\text{C}$ to $49^{\circ }\text{C}$ is

Imagine you just poured yourself a hot cup of tea. You leave it on the table and go answer a phone call. When you come back, it's cooler. But does it cool down at a constant speed? Not at all! The hotter the tea is compared to the room, the faster it loses heat. As it gets closer to room temperature, it gets lazy and cools down much slower. This everyday phenomenon is beautifully captured by Newton's Law of Cooling.

Strictly speaking, Newton's Law of Cooling is a differential equation. However, for small temperature drops (like $2^{\circ }\text{C}$ in our problem), we can use a highly accurate and much faster algebraic approximation known as the average form:

Here, $\frac{\Delta T}{\Delta t}$ is the rate of cooling, $K$ is a positive constant depending on the body and the surroundings, $T_{\text{avg}}$ is the average temperature of the body during the time interval, and $T_s$ is the constant temperature of the surroundings.

Let's break the problem into two distinct phases.

Phase 1: The Initial Rapid Cooling The body cools from $61^{\circ }\text{C}$ to $59^{\circ }\text{C}$ in $4\text{ min}$. - The temperature drop is $\Delta T=61-59=2^{\circ }\text{C}$. - The time taken is $\Delta t=4\text{ min}$. - The average temperature is $T_{\text{avg}}=\frac{61+59}{2}=60^{\circ }\text{C}$. - The surroundings are at $T_s=30^{\circ }\text{C}$.

Plugging these into our master equation:

Phase 2: The Slower Cooling Later, the body cools from $51^{\circ }\text{C}$ to $49^{\circ }\text{C}$ in an unknown time $t$. - The temperature drop is again $\Delta T=51-49=2^{\circ }\text{C}$. - The time taken is $\Delta t=t\text{ min}$. - The average temperature is now $T_{\text{avg}}=\frac{51+49}{2}=50^{\circ }\text{C}$.

Setting up the second equation:

We have a system of two equations. The constant $K$ is annoying, but we don't actually need to find its value! By simply dividing equation (i) by equation (ii), $K$ gracefully cancels out:

Now, it's just a matter of basic algebra to isolate $t$:

Notice the profound physical truth hidden in this simple number. It took $4\text{ min}$ to drop by $2^{\circ }\text{C}$ initially. But later, to drop by the exact same $2^{\circ }\text{C}$, it took $6\text{ min}$. Why? Because in the second phase, the body was at $50^{\circ }\text{C}$ (closer to the $30^{\circ }\text{C}$ room) compared to $60^{\circ }\text{C}$ in the first phase. A smaller temperature difference means a weaker driving force for heat transfer, resulting in a slower cooling rate.