Newton’s Law of Cooling is one of the most frequently tested topics in the CBSE Class 11 Physics syllabus, under the chapter Thermal Properties of Matter. It appears regularly in board exams as a conceptual question, a derivation, and a numerical problem. At Convex Classes, Jaipur, this concept is explained with complete clarity — definition, formula, derivation, graph, and solved examples — all in one place, so that students can revise efficiently and score full marks in this topic.

Definition

Newton’s Law of Cooling states that the rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings, provided this temperature difference is small.

In simpler terms: the hotter an object is compared to its surroundings, the faster it cools. As the object’s temperature approaches the surrounding temperature, its rate of cooling gradually decreases.

Everyday example: A freshly made cup of tea cools rapidly when it is very hot, but as it approaches room temperature, the rate of cooling slows down noticeably.

Formula

Differential Form

$$\frac{dT}{dt} = -k(T – T_s)$$dtdT​=−k(T−Ts​)

Symbol	Meaning
$T$T	Temperature of the body at time $t$t
$T_s$Ts​	Temperature of the surroundings
$t$t	Time
$k$k	Positive constant (cooling constant), depends on the nature of the surface, material, and surrounding medium
$\dfrac{dT}{dt}$dtdT​	Rate of change of temperature (rate of cooling)
Negative sign (−)	Indicates that temperature decreases with time

Integrated Form

$$T(t) = T_s + (T_0 – T_s)\,e^{-kt}$$T(t)=Ts​+(T0​−Ts​)e−kt

where $T_0$T0​ is the initial temperature of the body at $t = 0$t=0.

Approximate Form (most commonly used in numerical problems)

$$\frac{T_1 – T_2}{t} = k\left(\frac{T_1 + T_2}{2} – T_s\right)$$tT1​−T2​​=k(2T1​+T2​​−Ts​)

Here $T_1$T1​ is the initial temperature and $T_2$T2​ is the final temperature of the body over a given time interval $t$t.

Derivation (Step-by-Step)

This derivation is frequently asked directly in CBSE board exams for 3–4 marks.

Step 1: Starting from the statement of Newton’s Law of Cooling: $$\frac{dT}{dt} = -k(T – T_s)$$dtdT​=−k(T−Ts​)

Step 2: Separate the variables: $$\frac{dT}{T – T_s} = -k\, dt$$T−Ts​dT​=−kdt

Step 3: Integrate both sides: $$\int \frac{dT}{T – T_s} = -k \int dt$$∫T−Ts​dT​=−k∫dt$$\ln(T – T_s) = -kt + C$$ln(T−Ts​)=−kt+C

Step 4: Apply the initial condition — at $t = 0$t=0, $T = T_0$T=T0​: $$\ln(T_0 – T_s) = C$$ln(T0​−Ts​)=C

Step 5: Substitute the value of $C$C back into the equation: $$\ln(T – T_s) – \ln(T_0 – T_s) = -kt$$ln(T−Ts​)−ln(T0​−Ts​)=−kt$$\ln\left(\frac{T – T_s}{T_0 – T_s}\right) = -kt$$ln(T0​−Ts​T−Ts​​)=−kt

Step 6: Take the exponential of both sides: $$T – T_s = (T_0 – T_s)e^{-kt}$$T−Ts​=(T0​−Ts​)e−kt

Final Result:$$\boxed{T(t) = T_s + (T_0 – T_s)e^{-kt}}$$T(t)=Ts​+(T0​−Ts​)e−kt​

This confirms the integrated form of the law introduced in Section 3, with the complete derivation students should be able to reproduce in exams.

Graphical Representation

The cooling curve of Newton’s Law of Cooling is an exponential decay curve:

• The X-axis represents time ($t$t)
• The Y-axis represents temperature ($T$T)
• The curve starts at the initial temperature (T0T_0 T0​)
• It decreases exponentially with time
• The curve never falls below the surrounding temperature (TsT_s Ts​) — it approaches it asymptotically but never actually touches or crosses it

Key point to remember: Cooling is rapid initially (when the temperature difference is large) and becomes progressively slower as the body’s temperature approaches that of the surroundings. This is why the graph is a curve rather than a straight line.

Conditions and Assumptions

Newton’s Law of Cooling does not apply universally — the following conditions must hold:

1. The temperature difference between the body and its surroundings must be small (typically less than 20–30°C)
2. Heat loss must occur primarily through convection and radiation (conduction is assumed negligible)
3. The temperature of the surroundings must remain constant
4. The surface area of the body exposed to cooling must remain constant

Note: For larger temperature differences, Newton’s Law of Cooling does not give accurate results. In such cases, the Stefan-Boltzmann Law is used instead.

Solved Examples

Example 1 — Direct Application

Question: A cup of coffee is at 90°C. The room temperature is 25°C. If the cooling constant is $k = 0.05\, \text{min}^{-1}$k=0.05min−1, find the temperature of the coffee after 10 minutes.

Solution:$$T(t) = T_s + (T_0 – T_s)e^{-kt}$$T(t)=Ts​+(T0​−Ts​)e−kt$$T(10) = 25 + (90 – 25)e^{-0.05 \times 10}$$T(10)=25+(90−25)e−0.05×10$$T(10) = 25 + 65 \times e^{-0.5}$$T(10)=25+65×e−0.5$$T(10) = 25 + 65 \times 0.6065$$T(10)=25+65×0.6065$$T(10) = 25 + 39.4 = 64.4°C$$T(10)=25+39.4=64.4°C

Answer: The temperature of the coffee after 10 minutes will be approximately 64.4°C.

Example 2 — Using the Approximate Formula

Question: A body cools from 80°C to 60°C in 10 minutes. The surrounding temperature is 20°C. Find the cooling constant $k$k using Newton’s Law of Cooling.

Solution: Using the approximate formula: $$\frac{T_1 – T_2}{t} = k\left(\frac{T_1 + T_2}{2} – T_s\right)$$tT1​−T2​​=k(2T1​+T2​​−Ts​)$$\frac{80 – 60}{10} = k\left(\frac{80+60}{2} – 20\right)$$1080−60​=k(280+60​−20)$$2 = k(70 – 20)$$2=k(70−20)$$2 = k \times 50$$2=k×50$$k = 0.04\, \text{min}^{-1}$$k=0.04min−1

Answer: $k = 0.04\, \text{per minute}$k=0.04per minute

Example 3 — Conceptual Extension

If the same body from Example 2 continues to cool further, from 60°C to 40°C, the value of $k$k will remain the same, since $k$k is a constant for a given body and surrounding medium — it does not change with time or temperature. Recognising this is essential, as CBSE numericals frequently test this concept by providing data for two separate time intervals and expecting students to equate the two expressions for $k$k to find an unknown quantity (such as time or final temperature).

Applications

• Forensic Science: Estimating the time of death based on body temperature
• Culinary Science: Estimating how quickly hot food will cool
• Industrial Processes: Controlled cooling of metals after casting
• Meteorology: Building mathematical models of temperature variation
• Engineering: Understanding heat dissipation in engines and electronic devices

Common Mistakes to Avoid

1. Forgetting the negative sign in the differential formula
2. Assuming cooling is linear, when it is actually exponential
3. Applying the approximate formula to situations with a large temperature difference, where it is not valid
4. Treating the surrounding temperature ($T_s$Ts​) as a variable, when it is always constant in this law
5. Ignoring the units of $k$k, which are always the inverse of time (e.g., $\text{min}^{-1}$min−1 or $\text{s}^{-1}$s−1)

Quick Revision Summary

Point	Detail
Statement	Rate of cooling ∝ (Temperature of body − Temperature of surroundings)
Differential form	$\frac{dT}{dt} = -k(T – T_s)$dtdT​=−k(T−Ts​)
Integrated form	$T(t) = T_s + (T_0 – T_s)e^{-kt}$T(t)=Ts​+(T0​−Ts​)e−kt
Approximate form	$\frac{T_1-T_2}{t} = k\left(\frac{T_1+T_2}{2} – T_s\right)$tT1​−T2​​=k(2T1​+T2​​−Ts​)
Graph	Exponential decay curve
Valid for	Small temperature differences (convection + radiation only)
Applications	Forensics, cooking, industry, engineering

Frequently Asked Questions

Q1. In which class is Newton’s Law of Cooling taught in CBSE?

It is taught in Class 11 Physics, under the chapter Thermal Properties of Matter, and is also relevant for JEE and NEET preparation.

Q2. Does Newton’s Law of Cooling hold true for all temperature differences?

No, it is accurate only for small temperature differences. For larger differences, the Stefan-Boltzmann Law provides more accurate results.

Q3. What does the cooling constant kk k depend on?

It depends on the material, surface area, and nature of the surrounding medium (such as air or water).

Q4. Is the cooling curve linear or curved?

The cooling curve is exponential (curved), not linear — cooling is rapid initially and slows down over time.

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