State Newton's law of cooling and explain how it can be experimentally verified.

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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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Newton's law of cooling (§7.10) states that, for small temperature differences, the rate of loss of heat (equivalently, the rate of fall of temperature) of a body is directly proportional to the difference between its own temperature $T$ and that of its surroundings $T_0$ : $dT/dt \propto (T-T_0)$ , i.e. $dT/dt = -C(T-T_0)$ for some constant $C$ (Eq. 7.39). Experimental verification: a calorimeter is filled roughly two-thirds with boiling water, covered with a lid carrying a thermometer with its bulb immersed in the water, and left in surroundings whose own temperature stays roughly steady; the water's temperature is recorded once every minute as it falls, and plotted against time, giving a cooling curve (Fig. 7.13(a)) -- steep at first, gradually flattening as the water approaches room temperature. Tangents drawn to this curve at several points give the instantaneous rate of fall of temperature $dT/dt$ at each of t …

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$Q. Write Newton's law of Cooling.$
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Newton's law of cooling: −dQ/dt ∝ (T − T₀), i.e. the rate of cooling of a body is proportional to the excess of its temperature over the surroundings (for small temperature differences).

When a hot body is placed in cooler surroundings, it loses heat mainly by radiation and convection. Newton's law of cooling states that the rate of loss of heat, −dQ/dt, is directly proportional to the temperature difference (T − T₀) between the body (T) and its surroundings (T₀), as long as this difference is small: −dQ/dt = k(T − T₀), where k is a positive constant depending on the surface area and nature of the body. Since dQ = ms dT, this can also be written as −dT/dt = k'(T − …

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$Q. Newton's law of cooling is a special condition of which of the following? (A) Stefan's law (B) Boltzmann's law (C) Wien's law (D) Planck's law$
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Newton's law of cooling is a small-temperature-difference approximation of Stefan's law.

Stefan's (Stefan-Boltzmann) law gives the rate of radiant heat loss of a body at temperature $T$ in surroundings at $T_0$ as $\propto (T^4 - T_0^4)$ . When $T$ is only slightly greater than $T_0$ (i.e. $\Delta T = T - T_0$ is small), this can be expanded and approximated as directly proportional to $\Delta T$ :

…

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$Q. "The rate of heat-loss is proportional to the temperature difference of body and its surrounding." This is the statement of (A) Dalton's law (B) Stefan's law (C) Newton's law of cooling (D) Kirchhoff's law$
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This statement is Newton's law of cooling.

Newton's law of cooling states that, for a small temperature difference, the rate at which a body loses heat to its surroundings is directly proportional to the temperature difference between the body and its surroundings: $-\dfrac{dQ}{dt}\propto(T-T_0)$ . This is distinct from Stefan's law (radiat …

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$Q. "The rate of loss of heat -d(theta)/dt of the body is directly proportional to the temperature difference deltaT = (T2 - T1) of the body and surroundings." This statement is$
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This is precisely the statement of Newton's law of cooling, expressed mathematically as -dQ/dt is proportional to (T2 - T1).

Newton's law of cooling states that the rate at which a hot body loses heat to its surroundings is directly proportional to the temperature difference between the body and the surroundings, provided this difference is small:

-dQ/dt proportional to (T2 - T1) = ΔT

This is exactly the statement given in the question. It is distinct from: …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Newton's law of cooling is a special case of ?$
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Newton's law of cooling is a special case of Stefan's (Stefan–Boltzmann) law.

Stefan's law gives the net rate of radiation as $\propto (T^4 - T_0^4)$ . When the excess temperature (T − T₀) is small, expanding $T^4 - T_0^4$ and keeping the leading term makes the rate of cooling proportional to $(T - T_0)$ — which is exactly New …

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Q.State Newton's law of cooling and explain how it can be experimentally verified.

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