How to Derive and Apply Newton's Law of Cooling in Physics Problems
The Newton's law of cooling describes how the temperature of an object drops over time when it is warmer than its environment. For JEE Main, you need to understand both its mathematical form and the underlying concepts. This law is used to estimate cooling rates in practical physics scenarios, such as cooling tea or loss of heat from hot metals. JEE also tests you on its derivation, formula, and limitations—so mastering this topic strengthens your command over heat transfer chapters. Vedantu’s study modules cover such core concepts with exam-focussed clarity.
Newton's Law of Cooling: Definition, Key Statement and Real Context
Newton's law of cooling states that the rate at which a body loses heat is directly proportional to the temperature difference between the body and its surroundings, provided the difference is not too large. For example, if a heated metal is left on a table, it cools faster when much hotter than the room, and slower as it approaches room temperature.
- Applies when temperature difference is moderate (not extreme).
- Useful in studying thermal equilibrium processes.
- Limits: Not precise at very high or very low temperature gaps.
Newton's Law of Cooling Formula and Parameter Guide
The core equation for Newton's law of cooling is:
| Symbol | Meaning |
|---|---|
| T | Temperature of the object at time t (°C or K) |
| Ta | Ambient (surrounding) temperature (°C or K) |
| k | Cooling constant (depends on material, units: s-1) |
| t | Time elapsed (s) |
| T0 | Initial temperature at t = 0 |
The law is mathematically written as:
dT/dt = -k (T - Ta)
The negative sign means temperature falls over time. The constant k shows how quickly cooling happens, based on object’s properties and conditions.
Solving the differential equation gives:
T(t) = Ta + (T0 - Ta) e-kt
This exponential form often appears in JEE Main numericals. Try linking the use of exponential functions in physics for better understanding.
Stepwise Derivation of Newton's Law of Cooling
- Let T be object’s temperature, Ta ambient temperature.
- By law: Rate of cooling ∝ Temperature difference
- Write as: dT/dt = -k (T - Ta)
- Separate variables: dT/(T - Ta) = -k dt
- Integrate: ∫dT/(T-Ta) = ∫-k dt
- Logarithm: ln|T - Ta| = -kt + C
- At t = 0, T = T0: Find integration constant C.
- Solve for T: T(t) = Ta + (T0 - Ta) e-kt
This shows the temperature drops exponentially with time. Exam tip: always set up limits with clear initial values.
- Kinetic theory of gases links to thermal motion behind this law.
- Latent heat questions sometimes use cooling curves for calculations.
Graphical View: Understanding the Newton's Law of Cooling Curve
The Newton's law of cooling graph is an exponential cooling curve. The vertical axis is temperature T, horizontal is time t. The slope is steep at first (fast cooling) and gets flatter as T approaches Ta.
- Temperature never drops below ambient.
- Curve is steeper for higher k values.
- Used to estimate cooling times visually in exam questions.
Compare with the straight cooling assumed by simpler laws like first law of thermodynamics.
Applications, Limitations, and Traps in Newton's Law of Cooling
- Calculating time to cool food or drinks in daily life.
- Calorimetry experiments and studying cooling of liquids/solids in labs.
- Industrial use in designing cooling systems or loss calculations.
- Not valid for very large temperature differences or when overheating changes material properties.
- Fails when other heat transfer methods (like strong convection or phase change) dominate.
Common JEE mistakes include misusing the formula when T and Ta are close, confusing sign, or forgetting units of k (should be in s-1).
- Zeroth law of thermodynamics is a prerequisite for equilibrium conditions here.
- Contrast with Kirchhoff’s law for ideal emitter behaviour.
Solved Example: Applying Newton's Law of Cooling in JEE Context
A kettle of water cools from 80 °C to 60 °C in 10 minutes at room temperature 30 °C. If k is the cooling constant, how long will it take to reach 40 °C?
- Start: Use T(t) = Ta + (T0 - Ta) e-kt
- First cooling: 60 = 30 + (80-30) e-10k → e-10k = 30/50 = 0.6
- Second step: 40 = 30 + (80-30) e-tk → e-tk = 10/50 = 0.2
- Set up: (e-tk)/(e-10k) = 0.2/0.6 = 1/3 → e-k(t-10) = 1/3
- Take logs: -k(t-10) = ln(1/3)
- Solve for t: t = 10 - (ln(1/3)/k)
- From earlier, e-10k = 0.6 → -10k = ln(0.6) → k = -ln(0.6)/10 ≈ 0.051 s-1
- Therefore, t = 10 - [ln(1/3)/0.051] ≈ 10 + 21.56 ≈ 31.6 minutes
So, water will reach 40 °C after about 31.6 minutes.
- Practice more with thermodynamics mock test for deeper understanding.
- See laws of motion practice paper for application-focused problems.
Wrap-up: Mastery of Newton's Law of Cooling for JEE
By mastering the Newton's law of cooling, its calculus-based derivation, and recognizing its boundaries, you gain an advantage in JEE Main Physics. Combine it with topics like kinematics or properties of solids and liquids to solve integrated questions. Vedantu resources are great for topic-wise revision and practice. Remember, clear concept visualization and consistent formula application are your keys to scoring full marks in such heat transfer problems.
FAQs on Newton's Law of Cooling: Concept, Formula, Derivation & Examples
1. What is Newton's law of cooling?
Newton's law of cooling states that the rate at which an object cools is directly proportional to the difference between its temperature and the surrounding (ambient) temperature.
Key points:
- Applies when temperature difference is small
- A fundamental principle in thermodynamics and heat transfer
- Essential for exams like JEE and NEET
2. What is the formula for Newton’s law of cooling?
The formula for Newton’s law of cooling describes how temperature changes over time:
T(t) = T_{∞} + (T_0 - T_{∞}) ⋅ e-kt
Where:
- T(t) = temperature of object at time t
- T₀ = initial temperature of object
- T_{∞} = ambient temperature
- k = cooling constant (depends on material and conditions)
- t = time elapsed
3. How do you derive Newton's law of cooling?
Newton's law of cooling derivation involves calculus and key assumptions:
Stepwise derivation:
- Start with rate of cooling: −dT/dt ∝ (T - T_ambient)
- Express as differential equation: −dT/dt = k(T - T_ambient)
- Separate variables, integrate both sides
- Apply initial condition to solve for constant
- Obtain exponential form: T(t) = T_ambient + (T_initial - T_ambient) e^{-kt}
4. What is the cooling constant (k) in Newton's law of cooling?
The cooling constant (k) in Newton’s law of cooling measures how quickly an object cools.
Properties of k:
- Depends on nature of material, surface area, and type of environment
- Higher k = faster cooling
- Units: typically s⁻¹
- Determined experimentally or from graph slopes
5. What are three limitations of Newton’s law of cooling?
Three main limitations of Newton’s law of cooling are:
- Valid only for small temperature differences between body and ambient
- Assumes constant external conditions (like airflow and humidity)
- Not accurate for objects with variable heat capacity or changing phase
6. How to use Newton's law of cooling formula?
To use Newton’s law of cooling formula in numericals:
- Identify initial, final, and ambient temperatures and time interval
- Substitute values into T(t) = T_{∞} + (T_0 - T_{∞}) e^{-kt}
- Solve for the unknown (T, t, or k), often using logarithms
- Ensure consistency of units
7. Is Newton's law of cooling correct?
Newton's law of cooling is correct within its applicable limits.
- Best suited when temperature differences are moderate
- Loses accuracy at very high temperature differences or with complex heat transfer
- Widely used in Physics, Engineering, and Real-life cooling scenarios
8. Where is Newton’s law of cooling used in real life?
Newton’s law of cooling has several real-life applications:
- Designing cooling systems (radiators, refrigerators)
- Estimating time of death in forensics
- Engineering heat exchangers
- Predicting food/beverage cooling times
9. Does the law hold true for all temperature ranges?
No, Newton's law of cooling is reliable only for small to moderate temperature differences between the object and the environment. At very large temperature differences or with significant radiation, Stefan-Boltzmann law provides better accuracy.
10. Why does Newton’s law of cooling assume “small temperature differences”?
Newton’s law of cooling assumes small temperature differences because the proportionality between rate of heat loss and temperature difference holds true in this range.
- For large differences, other heat transfer effects (like radiation) become significant
- Ensures linear relationship as per the law
