Statement of the law
Place a hot body in surroundings held at a fixed temperature \(T_s\). The body loses heat and its temperature \(T\) falls. Newton's law of cooling states that the rate of loss of heat is directly proportional to the difference in temperature between the body and its surroundings, provided that difference is small. NCERT §10.10 writes the heat-rate form as
\[ -\frac{dQ}{dt} = k\,(T - T_s) \]where \(T - T_s\) is the excess temperature and \(k\) is a positive constant that depends on the exposed surface area and the nature of the surface. Since a body of mass \(m\) and specific heat \(s\) loses heat \(dQ = m s\, dT\), dividing through gives the more useful temperature-rate form:
\[ -\frac{dT}{dt} = K\,(T - T_s), \qquad K = \frac{k}{m s} \]Two readings of this equation must be kept apart. The proportionality is to the excess temperature \((T - T_s)\), never to the body's absolute temperature on its own. And the constant \(K\) is not a pure material property: it grows with surface area and shrinks with heat capacity \(m s\), so a thin sheet and a thick block of the same metal cool at different rates.
| Symbol | Meaning | Note |
|---|---|---|
| \(T\) | Instantaneous temperature of the body | Falls with time toward \(T_s\) |
| \(T_s\) | Temperature of the surroundings | Held constant; the floor the curve approaches |
| \(T - T_s\) | Excess temperature | The driver of cooling; rate \(\propto\) this |
| \(k\) | Heat-rate constant in \(-dQ/dt = k(T-T_s)\) | Depends on area and surface nature |
| \(K = k/ms\) | Cooling constant in \(-dT/dt = K(T-T_s)\) | Slope of the \(\ln(T-T_s)\) line |
Why it is an approximation of Stefan's law
Newton's law is not a fundamental law. It is the small-difference limit of Stefan's law of thermal radiation, and that is exactly why it carries the "small temperature difference" caveat. A body at temperature \(T\) in surroundings at \(T_s\) radiates a net power per unit area
\[ E = e\sigma\,(T^4 - T_s^4)\,A \]Factorising the difference of fourth powers as a difference of squares twice gives
\[ T^4 - T_s^4 = (T - T_s)(T^3 + T^2 T_s + T T_s^2 + T_s^3). \]When the excess \((T - T_s)\) is small, \(T\approx T_s\), so each of the four terms in the second bracket is close to \(T_s^3\) and the bracket reduces to \(4T_s^3\). The net loss then collapses to
\[ E \approx e\sigma\,(4 T_s^3 A)\,(T - T_s) = k\,(T - T_s), \qquad k = 4 e \sigma T_s^3 A. \]This is Newton's law, recovered as a linear approximation. The argument also makes the limit precise: it holds while \((T - T_s)\) stays small enough that the higher powers of the excess can be dropped. For large excess temperatures the genuine \(T^4\) behaviour dominates and the linear law fails.
The exponential cooling solution
The temperature-rate form is a first-order differential equation. Separate the variables and integrate from the initial temperature \(T_0\) at \(t=0\):
\[ \frac{dT}{T - T_s} = -K\,dt \quad\Longrightarrow\quad \ln(T - T_s) = -Kt + c. \]Exponentiating, with the constant fixed by \(T = T_0\) at \(t=0\), gives the cooling law in closed form:
\[ \boxed{\,T = T_s + (T_0 - T_s)\,e^{-Kt}\,} \]The body's temperature decays exponentially from \(T_0\) toward the surrounding temperature \(T_s\). At \(t=0\) the excess is \((T_0 - T_s)\); it shrinks by a factor \(e^{-Kt}\), so the curve is steepest at the start and flattens as it approaches \(T_s\). The body never quite reaches \(T_s\) in finite time, and the curve never dips below it.
The straight-line ln-graph method
The exponential curve is hard to test against by eye — many decaying curves look similar. The clean experimental check is to take logarithms. The integrated form
\[ \ln(T - T_s) = -Kt + c \]is the equation of a straight line in the variables \(\ln(T - T_s)\) and \(t\). Its slope is \(-K\) and its intercept is \(\ln(T_0 - T_s)\). A genuinely straight plot is the signature that the data obey Newton's law, and the cooling constant \(K\) is read directly off the negative slope.
The average-temperature method
For a body cooling through a small step from \(T_1\) to \(T_2\) over a time \(\Delta t\), the instantaneous rate can be replaced by the finite ratio \((T_1 - T_2)/\Delta t\), and the excess temperature is evaluated at the mean of the two endpoints. This gives the working form NEET problems use most:
\[ \frac{T_1 - T_2}{\Delta t} = K\!\left(\frac{T_1 + T_2}{2} - T_s\right) \]The power of this form is that the same \(K\) governs every stage of cooling for a given body. Writing the equation for two different temperature intervals and dividing eliminates \(K\) entirely, so a "how long to cool from A to B" question reduces to a single ratio. This is precisely the structure of the recurring NEET coffee-and-tea problems.
Newton's law follows from radiation physics — revisit heat transfer modes for Stefan's law and the \(T^4\) dependence it approximates.
Experimental verification
NCERT describes a direct laboratory check using a double-walled vessel (Fig. 10.20). The procedure is short and worth memorising as a sequence.
| Step | Action | Purpose |
|---|---|---|
| 1 | Fill a copper calorimeter with hot water and place it inside a double-walled vessel holding water at a steady temperature. | Keeps the surroundings \(T_s\) effectively constant. |
| 2 | Read the calorimeter temperature \(T\) at equal time intervals using a thermometer, stirring gently. | Gives the \(T\)-versus-\(t\) data. |
| 3 | Plot \(\ln(T - T_s)\) against time \(t\). | Linearises the exponential decay. |
| 4 | Confirm the plot is a straight line with negative slope. | Verifies the law; slope magnitude equals \(K\). |
The observed straight line with negative slope is the experimental support for the integrated equation \(\ln(T-T_s) = -Kt + c\). NCERT notes that for small differences the combined loss by conduction, convection and radiation is together proportional to the excess temperature, which is why the law applies well to everyday cases such as a radiator warming a room or a cup of tea cooling on a table.
Worked example
A pan filled with hot food cools from 94 °C to 86 °C in 2 minutes when the room temperature is 20 °C. How long will it take to cool from 71 °C to 69 °C?
First interval. The mean of 94 °C and 86 °C is 90 °C, which is \(90 - 20 = 70\) °C above the room. The pan drops 8 °C in 2 minutes, so the average form gives \(\dfrac{8\,^\circ\text{C}}{2\,\text{min}} = K\,(70\,^\circ\text{C})\).
Second interval. The mean of 71 °C and 69 °C is 70 °C, which is \(70 - 20 = 50\) °C above the room. The pan drops 2 °C in an unknown time, so \(\dfrac{2\,^\circ\text{C}}{\text{Time}} = K\,(50\,^\circ\text{C})\).
Divide to cancel \(K\). Dividing the first equation by the second, \(\dfrac{8/2}{2/\text{Time}} = \dfrac{70}{50}\). Rearranging gives \(\text{Time} = \dfrac{2 \times 2 \times 50}{8 \times 70} = 0.7\) min \(= 42\) s.
Answer: the pan takes about 42 seconds to cool from 71 °C to 69 °C — far less than 2 minutes, because its excess temperature is now smaller, so it cools more slowly per degree yet drops a smaller range.
Newton's law of cooling in one breath
- Rate of heat loss \(-\dfrac{dQ}{dt} = k(T - T_s)\); temperature form \(-\dfrac{dT}{dt} = K(T - T_s)\) with \(K = k/ms\).
- Proportional to the excess temperature \((T - T_s)\), never to the absolute temperature \(T\) alone.
- Valid only for small \((T - T_s)\): it is the linear approximation of Stefan's \(T^4 - T_s^4\) law, with \(k = 4e\sigma T_s^3 A\).
- Solution is exponential decay: \(T = T_s + (T_0 - T_s)e^{-Kt}\); steepest at first, asymptotic to \(T_s\).
- Plot \(\ln(T - T_s)\) versus \(t\) for a straight line of slope \(-K\) — the verification graph.
- Average-temperature form \(\dfrac{T_1 - T_2}{\Delta t} = K\!\left(\dfrac{T_1+T_2}{2} - T_s\right)\) solves the cooling-from-A-to-B PYQs.