What Newton’s Law of Cooling Describes
When an object is warmer than its surroundings, it generally loses heat; when it is cooler, it generally gains heat. Newton’s law of cooling models the rate of temperature change as proportional to the difference between the object’s temperature and the surrounding temperature.
The model is useful when the ambient temperature stays approximately constant and the object’s cooling behavior can be represented by one constant rate parameter. It leads to an exponential change in the temperature difference—not, in general, an exponential change in the object’s temperature itself.
The sign of the rate follows from the temperature difference. If \(T>T_a\), then \(\frac{dT}{dt}<0\), so the object cools. If \(T<T_a\), then \(\frac{dT}{dt}>0\), so it warms. If \(T=T_a\), then the rate is zero.
Deriving the Temperature Model
Suppose the object begins at temperature \(T_0\) when \(t=0\). If \(T_0\ne T_a\), separate the variables in Newton’s law:
Integrating both sides gives
Exponentiating shows that the temperature difference has the form \(T-T_a=Ae^{-kt}\), where \(A\) is a constant that can be positive or negative. Applying \(T(0)=T_0\) gives \(A=T_0-T_a\). Thus the solution is
If \(T_0=T_a\), dividing by \(T-T_a\) is not allowed. In that case the constant function \(T(t)=T_a\) directly satisfies both the differential equation and the initial condition. The displayed formula also gives this constant solution when \(T_0=T_a\).
The derivation can be checked by differentiating the proposed solution. Its derivative is \(-k(T_0-T_a)e^{-kt}\), while its difference from ambient temperature is \((T_0-T_a)e^{-kt}\). Therefore its derivative equals \(-k(T-T_a)\), as required. At \(t=0\), it gives \(T(0)=T_a+(T_0-T_a)=T_0\).
Worked Example: Finding the Temperature After a Set Time
Worked Example: Finding the Temperature After a Set Time
An object is initially at \(80^\circ\text{C}\) in a room held at \(20^\circ\text{C}\). The cooling constant is \(k=0.10\) per minute. Find its temperature after 10 minutes.
Here \(T_a=20\), \(T_0=80\), and \(k=0.10\). Substitute these values into the solution:
Using \(e^{-1}\approx0.367879\),
Rounded to one decimal place, the predicted temperature is \(42.1^\circ\text{C}\). This lies between the initial temperature and the ambient temperature, as expected. To check the model, differentiating \(T(t)=20+60e^{-0.10t}\) gives \(T'(t)=-6e^{-0.10t}\). Also, \(-k(T-20)=-0.10(60e^{-0.10t})=-6e^{-0.10t}\), so the function satisfies the differential equation and has \(T(0)=80\).
Finding the Cooling Constant from Measurements
Temperature measurements can be used to determine \(k\). If the ambient temperature is known, and the object’s temperatures \(T_1\) and \(T_2\) are recorded at times \(t_1\) and \(t_2\), the solution gives
Dividing the second equation by the first cancels the initial temperature difference. For \(t_2>t_1\), this gives
This calculation requires that both temperature differences be nonzero and that their ratio be positive. Under this model, an object that has not reached ambient temperature stays on the same side of ambient temperature, so the differences have the same sign. For \(k>0\), the later difference is smaller in magnitude, making the logarithm in the expression for \(k\) positive.
Worked Example: Estimating \(k\) and Predicting a Later Temperature
A \(80^\circ\text{C}\) object is placed in surroundings at \(20^\circ\text{C}\). After 5 minutes, its temperature is \(50^\circ\text{C}\). Find \(k\), then predict its temperature after 15 minutes.
The temperature difference from ambient falls from \(80-20=60^\circ\text{C}\) to \(50-20=30^\circ\text{C}\) in 5 minutes. The solution gives
Divide by 60 and take natural logarithms:
At \(t=15\) minutes,
Check against the measurement at 5 minutes: the model gives \(T(5)=20+60e^{-\ln 2}=20+30=50^\circ\text{C}\), as recorded. The rate constant is positive, and the predicted temperature remains above ambient, consistent with the situation.
Using Two Measurements That Are Both After the Start
The initial temperature does not need to be known if two later measurements and the ambient temperature are available. The quotient formula uses only those readings and the time between them.
Worked Example: Using Two Later Temperature Readings
An object is in a room at \(18^\circ\text{C}\). Its temperature is \(58^\circ\text{C}\) at \(t=4\) minutes and \(28^\circ\text{C}\) at \(t=10\) minutes. Estimate \(k\), then predict its temperature at \(t=13\) minutes.
Relative to ambient, the measured differences are \(58-18=40^\circ\text{C}\) and \(28-18=10^\circ\text{C}\). There are \(10-4=6\) minutes between readings, so
Taking logarithms gives \(\ln(1/4)=-6k\), so
From \(t=10\) to \(t=13\) is 3 minutes. The temperature difference at \(t=10\) is \(10^\circ\text{C}\), so the predicted difference at \(t=13\) is \(10e^{-3k}\):
Check the model against both readings. From \(t=4\) to \(t=10\), the difference changes from \(40\) to \(40e^{-6(\ln 2/3)}=40e^{-2\ln 2}=40/4=10\), matching the \(28^\circ\text{C}\) reading. The initial reading at \(t=4\) is \(18+40=58^\circ\text{C}\), also matching. The later prediction is above ambient and closer to it than either measured temperature.
Why the Temperature Difference Matters
The equation is proportional to \(T-T_a\), not to \(T\). The ambient temperature is the reference level toward which the object moves. The solution makes this especially clear:
Because \(e^{-kt}\) is positive and decreases toward zero when \(k>0\), the temperature difference does not change sign and its magnitude shrinks. The object approaches ambient temperature but does not cross it in this constant-parameter model. This describes the mathematical behavior of the model; real measurements may differ if ambient conditions or the object’s cooling behavior change.
Units also provide a useful check. If time is measured in minutes, then \(kt\) must be dimensionless, so \(k\) must have units of \(\text{min}^{-1}\). In the exponential, \(e^{-kt}\), the exponent is therefore unit-free. A temperature difference has the same temperature units as \(T\) and \(T_a\), so the formula adds quantities with compatible units.
Common Mistake: Losing the Negative Sign
From the measurement in the second example, \(1/2=e^{-5k}\). A common error is to write \(k=\ln(1/2)/5\), which is negative because \(\ln(1/2)<0\). That contradicts the assumption \(k>0\) and would make \(e^{-kt}\) grow rather than decay.
The correct logarithm equation is \(\ln(1/2)=-5k\). Dividing by \(-5\) gives
A practical check is that a cooling constant found from a later reading should be positive, and the predicted temperature difference should decrease in magnitude over time.
Check Your Understanding
Use the model and reasoning developed above to answer these questions.
- An object begins at \(70^\circ\text{C}\) in a \(22^\circ\text{C}\) room. Write its temperature function in terms of \(k\).
- Explain why an object initially cooler than its surroundings has a positive temperature rate under Newton’s law.
- An object’s temperature difference from ambient falls from \(48^\circ\text{C}\) to \(24^\circ\text{C}\) in 8 minutes. Find \(k\) in inverse minutes.
- What does the model predict when the initial temperature equals ambient temperature?
- Why must the time units used with \(k\) agree with the time units used in the exponential model?
