A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correc

NCERT Exemplar · Q12

Q. A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct? (Note: more than one of the given options may be correct.)

(a) The rate of cooling is constant till milk attains the temperature of the surrounding.

(b) The temperature of milk falls off exponentially with time.

(c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surounding and that is why it cools.

(d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings.

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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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Cooling of a hot liquid is governed by Newton’s law of cooling, which gives an exponential temperature fall only under ideal conditions; the net heat flow is always from hot to cold, and all three heat transfer modes (conduction, convection, radiation) contribute. The correct options are (B), (C), and (D).

Let’s examine each statement carefully, keeping the physics of thermal radiation and heat transfer in mind.

Option (A): “The rate of cooling is constant till milk attains the temperature of the surrounding.” Newton’s law of cooling states that the rate of heat loss is proportional to the temperature difference between the object and its surroundings:

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

As the milk cools, $T$ decreases, so the temperature difference shrinks. Hence the rate of cooling itself decreases — it is not constant. Only if the temperature difference were artificially kept constant (e.g., by a heater) would the rate be constant. So (A) is false.

Option (B): “The temperature of milk falls off exponentially with time.” From Newton’s law, if the temperature difference is small and the heat capacity is constant, we get:

\frac{dT}{dt} = -k (T - T_{\text{surr}})

Solving this differential equation gives:

T(t) = T_{\text{surr}} + (T_0 - T_{\text{surr}}) e^{-kt}

This is an exponential decay toward the surrounding temperature. However, this is an approximation valid when the temperature difference is not too large and radiative losses are linearised. For a cup of hot milk, the difference is moderate, so the exponential model works well. Hence (B) is correct.

Watch out

The exponential law is exact only when the cooling is purely convective with constant coefficients. At very high temperatures, radiation follows a $T^4$ law, making the fall steeper than exponential initially. But for typical milk temperatures (≈60–70°C), the exponential approximation is excellent.

Option (C): “While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools.” …

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Q.A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct? (Note: more than one of the given options may be correct.)

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