Q.A body cools according to Newton's law from $100^\circ$c to $60^\circ$c in 20 minutes, with the surroundings at $20^\circ$c. How long will it take to cool down to $30^\circ$c?

Newton's Law of Cooling: From Intuition to Formula

Imagine you pour a cup of hot coffee. You know it will cool down, but how fast? If the coffee is scalding hot, it cools quickly at first. As it gets closer to room temperature, the cooling slows down — it takes much longer to go from 40°C to 30°C than from 90°C to 80°C. That's the core observation.

The intuition: The hotter an object is relative to its surroundings, the faster it loses heat. The driving force for cooling is the temperature difference between the object and the environment. When that difference is large, heat rushes out. When the difference is small, heat trickles out.

The Precise Statement

Newton's Law of Cooling states:

The rate of heat loss of a body is directly proportional to the difference in temperature between the body and its surroundings, provided the temperature difference is small and the mode of heat transfer is primarily convection (and radiation, in some cases).

Let's break that down.

Mathematically:

If $T(t)$ is the temperature of the object at time $t$, and $T_s$ is the constant temperature of the surroundings (the "ambient" temperature), then:

The negative sign is crucial: it tells us the temperature decreases when $T > T_s$ (cooling) and increases when $T < T_s$ (warming — the law works for heating too).

Introducing a positive constant $k$ (which depends on the object's surface area, material, and the surrounding medium), we get the differential equation:

This is a simple first-order differential equation. Its solution, which gives the temperature at any time, is:

where $T_0$ is the initial temperature of the object at $t = 0$.

What the Solution Tells You

• Exponential decay of the temperature difference. The quantity $(T - T_s)$ shrinks exponentially toward zero. The object never exactly reaches $T_s$ in finite time, but it gets arbitrarily close.
• The constant $k$ controls the speed. A larger $k$ means faster cooling (e.g., a thin metal cup vs. a thick ceramic mug). A smaller $k$ means slower cooling.
• The surroundings temperature $T_s$ is the asymptote. The object's temperature approaches $T_s$ from above (cooling) or below (heating).

A Quick Example

A cup of tea at 90°C is placed in a room at 20°C. After 5 minutes, it's 60°C. Find the temperature after another 5 minutes.

Step 1: Identify $T_0 = 90$, $T_s = 20$, $t = 5$ min, $T(5) = 60$.

From the solution: $60 = 20 + (90 - 20) e^{-5k}$ → $40 = 70 e^{-5k}$ → $e^{-5k} = \frac{4}{7}$ → $k = -\frac{1}{5} \ln\left(\frac{4}{7}\right) \approx 0.112$ per minute.

Step 2: Find $T(10)$: $T(10) = 20 + 70 e^{-10k} = 20 + 70 \left(e^{-5k}\right)^2 = 20 + 70 \left(\frac{4}{7}\right)^2 = 20 + 70 \cdot \frac{16}{49} = 20 + \frac{1120}{49} \approx 42.86^\circ\text{C}$.

Notice: in the first 5 minutes, it dropped 30°C. In the next 5 minutes, it dropped only about 17°C. That's the law in action.

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