One day in the morning, Ramesh filled up 1/3 bucket of hot water from geyser, to take bath. Remaining 2/3 was

NCERT Exemplar · Q21

Q. One day in the morning, Ramesh filled up 1/3 bucket of hot water from geyser, to take bath. Remaining 2/3 was to be filled by cold water (at room temperature) to bring mixture to a comfortable temperature. Suddenly Ramesh had to attend to something which would take some times, say 5-10 minutes before he could take bath. Now he had two options:

(i) fill the remaining bucket completely by cold water and then attend to the work,

(ii) first attend to the work and fill the remaining bucket just before taking bath. Which option do you think would have kept water warmer? Explain.

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Concept understanding — Newton's Law of Cooling

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:

\frac{dT}{dt} \propto -(T - T_s)

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:

\frac{dT}{dt} = -k (T - T_s)

\frac{dT}{dt} = -k (T - T_s)

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

T(t) = T_s + (T_0 - T_s) e^{-kt}

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).

Note

The law is an approximation. It works well for moderate temperature differences (say, up to a few tens of degrees) and when the surroundings are large enough that $T_s$ stays constant. For very large differences (e.g., a red-hot iron in air), radiation becomes dominant and the law breaks down.

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.

Common Mistakes to Avoid …

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To keep the water warmer, it is better to mix the hot and cold water immediately. This reduces the overall temperature difference with the surroundings, thereby slowing down the rate of heat loss during the waiting period.

The core concept here is Newton's Law of Cooling, which states that the rate of heat loss from a body is approximately proportional to the temperature difference between the body and its surroundings.

\frac{dQ}{dt} \propto (T - T_{surr})

Where $T$ is the temperature of the body and $T_{surr}$ is the temperature of the surroundings. This means that the hotter an object is relative to its surroundings, the faster it will lose heat.

Let's analyze both options based on this principle:

Option (i): Fill the remaining bucket completely by cold water and then attend to the work.
- Initial State: Ramesh has $1/3$ bucket of hot water at temperature $T_H$ and $2/3$ bucket of cold water at room temperature $T_C$ .
- Action: He immediately mixes the hot and cold water. Let the total mass of water be $M$ . So, the mass of hot water is $m_H = M/3$ and the mass of cold water is $m_C = 2M/3$ .
- Mixing: When mixed, the hot water loses heat and the cold water gains heat until they reach a common equilibrium temperature, $T_{mix}$ . Assuming no heat loss to the surroundings during this quick mixing process:

m_H c (T_H - T_{mix}) = m_C c (T_{mix} - T_C)

\frac{M}{3} c (T_H - T_{mix}) = \frac{2M}{3} c (T_{mix} - T_C)

T_H - T_{mix} = 2T_{mix} - 2T_C

3T_{mix} = T_H + 2T_C

T_{mix} = \frac{T_H + 2T_C}{3}

*   **Waiting Period:** Now, the entire bucket of water (mass $M$) is at the temperature $T_{mix}$. This mixture then sits for 5-10 minutes, losing heat to the surroundings (which are at $T_C$).
*   **Heat Loss Rate:** The rate of heat loss from this mixture will be proportional to $(T_{mix} - T_C)$.

T_{mix} - T_C = \frac{T_H + 2T_C}{3} - T_C = \frac{T_H + 2T_C - 3T_C}{3} = \frac{T_H - T_C}{3}

    So, the temperature difference driving heat loss is $\frac{T_H - T_C}{3}$.

2. Option (ii): First attend to the work and fill the remaining bucket just before taking bath.

* Initial State: Ramesh has $1/3$ bucket of hot water at temperature $T_H$ . The remaining $2/3$ is empty.

* Waiting Period: This $1/3$ bucket of hot water (mass $m_H = M/3$ ) sits for 5-10 minutes, losing heat to the surroundings (at $T_C$ ).

* Heat Loss Rate: During this period, the hot water is at a high temperature $T_H$ . The rate of heat loss will be proportional to $(T_H - T_C)$ .

* Comparison of Heat Loss Rates: Notice that $(T_H - T_C)$ in Option (ii) is significantly larger than $\frac{T_H - T_C}{3}$ in Option (i). This means the hot water in Option (ii) will lose heat at a much faster rate during the waiting period. …

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