Introduction
Suppose you place 200 ml of water in a pot on a stove and heat it until it boils.
At first, this sounds like a simple everyday situation. But mathematically, it is a beautiful example of how differential equations are used to model real systems.
The main question is:
How does the temperature of the water change over time?
A differential equation helps us describe that change.
In this blog, we will carefully formulate the problem, identify the assumptions, build the differential equation, solve it, and interpret the result.
The Physical Situation
We place:
$$200\text{ ml}$$
of water on a stove.
Since the density of water is approximately:
$$1\text{ g/ml}$$
the mass is approximately:
$$m=200\text{ g}=0.2\text{ kg}$$
Suppose the water starts at room temperature:
$$T(0)=20^\circ C$$
We want to know how long it takes to reach boiling temperature:
$$100^\circ C$$
assuming normal atmospheric pressure.
Step 1: Identify the Quantity That Changes
The main variable is the water temperature.
Let:
$$T(t)=\text{temperature of water at time }t$$
where:
- $$T(t)$$ is measured in degrees Celsius
- $$t$$ is measured in seconds
- $$T(0)$$ is the initial temperature
We want to model:
$$\frac{dT}{dt}$$
This means:
the rate at which the water temperature changes over time.
Step 2: Use an Energy Balance
The basic physics principle is:
Energy entering the water increases the thermal energy of the water.
The thermal energy needed to raise water temperature is:
$$Q=mc\Delta T$$
where:
- $$Q$$ = heat energy in joules
- $$m$$ = mass of water
- $$c$$ = specific heat capacity of water
- $$\Delta T$$ = change in temperature
For water:
$$c\approx4186\text{ J/kg}^\circ C$$
The rate form is:
$$\frac{dQ}{dt}=mc\frac{dT}{dt}$$
The term:
$$\frac{dQ}{dt}$$
is power, measured in watts.
So:
$$\text{Power into water}=mc\frac{dT}{dt}$$
Step 3: Start With the Simplest Model
The simplest assumption is:
The stove transfers a constant effective power into the water.
Let:
$$P=\text{effective power absorbed by the water}$$
Then:
$$mc\frac{dT}{dt}=P$$
So:
$$\frac{dT}{dt}=\frac{P}{mc}$$
This is our first differential equation.
Step 4: Solve the Simple Model
We have:
$$\frac{dT}{dt}=\frac{P}{mc}$$
Since the right side is constant, integrate both sides:
$$T(t)=T(0)+\frac{P}{mc}t$$
This says temperature rises linearly over time.
Step 5: Estimate the Time to Boil
We want:
$$T(t)=100$$
So:
$$100=T(0)+\frac{P}{mc}t$$
Rearrange:
$$t=\frac{mc(100-T(0))}{P}$$
Using:
$$m=0.2$$
$$c=4186$$
$$T(0)=20$$
we get:
$$t=\frac{0.2(4186)(100-20)}{P}$$
Calculate the energy needed:
$$Q=0.2(4186)(80)$$
$$Q=66976\text{ J}$$
So the time is:
$$t=\frac{66976}{P}$$
If the water absorbs:
$$P=500\text{ W}$$
then:
$$t=\frac{66976}{500}=133.95\text{ seconds}$$
That is about:
$$2.23\text{ minutes}$$
This is an idealized estimate.
Why This Simple Model Is Too Simple
In real life, the temperature usually does not rise perfectly linearly.
Why?
Because some heat is lost to the surroundings.
Heat is lost through:
- the pot
- the air
- evaporation
- radiation
- convection
- heating the pot itself
As the water gets hotter, heat loss increases.
So we need a better model.
Step 6: Add Heat Loss
A common assumption is Newton’s Law of Cooling.
It says heat loss is proportional to the temperature difference between the water and the room.
Let:
$$T_a=\text{ambient room temperature}$$
and let:
$$h=\text{heat loss coefficient}$$
Then heat loss is:
$$h(T-T_a)$$
Now the energy balance becomes:
$$mc\frac{dT}{dt}=P-h(T-T_a)$$
This is a more realistic differential equation.
Step 7: Interpret the Differential Equation
The equation is:
$$mc\frac{dT}{dt}=P-h(T-T_a)$$
The left side is:
$$mc\frac{dT}{dt}$$
This represents the rate at which the water’s stored thermal energy is changing.
The right side has two parts.
First:
$$P$$
is heat entering from the stove.
Second:
$$h(T-T_a)$$
is heat escaping to the environment.
So the equation says:
Temperature increases when stove input exceeds heat loss.
If the water is cool, heat loss is small.
If the water is hot, heat loss is larger.
Step 8: Put the Equation in Standard Form
Starting with:
$$mc\frac{dT}{dt}=P-h(T-T_a)$$
Divide by:
$$mc$$
This gives:
$$\frac{dT}{dt}=\frac{P}{mc}-\frac{h}{mc}(T-T_a)$$
Let:
$$k=\frac{h}{mc}$$
Then:
$$\frac{dT}{dt}=\frac{P}{mc}-k(T-T_a)$$
This is a first-order linear differential equation.
Step 9: Find the Equilibrium Temperature
At equilibrium:
$$\frac{dT}{dt}=0$$
So:
$$0=\frac{P}{mc}-k(T-T_a)$$
Since:
$$k=\frac{h}{mc}$$
we can also write:
$$0=P-h(T-T_a)$$
Therefore:
$$T=T_a+\frac{P}{h}$$
This value is called the steady-state temperature.
Let:
$$T_\infty=T_a+\frac{P}{h}$$
This is the temperature the system would approach if heating continued forever and boiling did not intervene.
Step 10: Solve the Differential Equation
The equation can be written as:
$$\frac{dT}{dt}=k(T_\infty-T)$$
where:
$$T_\infty=T_a+\frac{P}{h}$$
This means:
The rate of heating is proportional to the gap between the current temperature and the limiting temperature.
Now solve:
$$\frac{dT}{dt}=k(T_\infty-T)$$
Separate variables:
$$\frac{dT}{T_\infty-T}=kdt$$
Integrate both sides:
$$-\ln(T_\infty-T)=kt+C$$
Rearrange:
$$T_\infty-T=Ce^{-kt}$$
So:
$$T(t)=T_\infty-Ce^{-kt}$$
Use the initial condition:
$$T(0)=T_0$$
Then:
$$T_0=T_\infty-C$$
So:
$$C=T_\infty-T_0$$
Therefore:
$$T(t)=T_\infty-(T_\infty-T_0)e^{-kt}$$
This is the solution.
Step 11: Interpretation of the Solution
The solution is:
$$T(t)=T_\infty-(T_\infty-T_0)e^{-kt}$$
where:
- $$T_0$$ = starting temperature
- $$T_\infty$$ = limiting temperature
- $$k$$ = heating constant
At the beginning:
$$t=0$$
So:
$$e^{-k(0)}=1$$
Then:
$$T(0)=T_\infty-(T_\infty-T_0)=T_0$$
As time increases:
$$e^{-kt}\rightarrow0$$
So:
$$T(t)\rightarrow T_\infty$$
The water temperature approaches the limiting temperature.
Step 12: Time to Reach Boiling
We want to find when:
$$T(t)=100$$
Using:
$$T(t)=T_\infty-(T_\infty-T_0)e^{-kt}$$
Set:
$$100=T_\infty-(T_\infty-T_0)e^{-kt}$$
Rearrange:
$$(T_\infty-T_0)e^{-kt}=T_\infty-100$$
Divide:
$$e^{-kt}=\frac{T_\infty-100}{T_\infty-T_0}$$
Take natural logarithms:
$$-kt=\ln\left(\frac{T_\infty-100}{T_\infty-T_0}\right)$$
Therefore:
$$t=-\frac{1}{k}\ln\left(\frac{T_\infty-100}{T_\infty-T_0}\right)$$
This gives the time required to reach boiling.
Step 13: A Numerical Example
Suppose:
$$T_0=20^\circ C$$
$$T_a=20^\circ C$$
$$m=0.2\text{ kg}$$
$$c=4186\text{ J/kg}^\circ C$$
Suppose the effective power into the water is:
$$P=500\text{ W}$$
Assume heat loss coefficient:
$$h=2\text{ W}/^\circ C$$
Then:
$$T_\infty=T_a+\frac{P}{h}=20+\frac{500}{2}=270^\circ C$$
and:
$$k=\frac{h}{mc}=\frac{2}{0.2(4186)}$$
$$k\approx0.00239$$
Now calculate time to boil:
$$t=-\frac{1}{0.00239}\ln\left(\frac{270-100}{270-20}\right)$$
$$t=-418.4\ln\left(\frac{170}{250}\right)$$
$$t=-418.4\ln(0.68)$$
$$t\approx161.2\text{ seconds}$$
So the estimated boiling time is about:
$$2.7\text{ minutes}$$
This is slightly slower than the ideal model because heat loss was included.
Step 14: What Happens at Boiling?
Until the water reaches:
$$100^\circ C$$
the heat mainly increases temperature.
But once water reaches boiling point, the temperature stops increasing.
Additional heat goes into changing liquid water into steam.
This is called latent heat of vaporization.
For water:
$$L_v\approx2260\text{ kJ/kg}$$
If all 200 ml of water were boiled away, the energy needed would be:
$$Q=mL_v$$
$$Q=0.2(2260000)$$
$$Q=452000\text{ J}$$
This is much larger than the energy needed to heat water from 20°C to 100°C.
Heating to boiling required about:
$$66976\text{ J}$$
Boiling all the water away requires about:
$$452000\text{ J}$$
So reaching boiling is much easier than evaporating all the water.
Step 15: A Piecewise Model
A better model has two stages.
Stage 1: Heating Phase
For:
$$T(t)<100$$
we use:
$$mc\frac{dT}{dt}=P-h(T-T_a)$$
Stage 2: Boiling Phase
Once:
$$T(t)=100$$
the temperature remains approximately constant:
$$T(t)=100$$
and the heat goes into vaporization.
The mass begins to decrease:
$$L_v\frac{dm}{dt}=-(P-h(100-T_a))$$
So:
$$\frac{dm}{dt}=-\frac{P-h(100-T_a)}{L_v}$$
This models the water boiling away.
Step 16: How to Think Like a Modeler
This example teaches an important modeling process.
1. Identify the state variable
The main variable is:
$$T(t)$$
the water temperature.
2. Identify inputs
The stove supplies power:
$$P$$
3. Identify losses
Heat escapes at rate:
$$h(T-T_a)$$
4. Write a balance law
Energy balance gives:
$$mc\frac{dT}{dt}=P-h(T-T_a)$$
5. Solve the equation
This gives:
$$T(t)=T_\infty-(T_\infty-T_0)e^{-kt}$$
6. Interpret the result
The water warms quickly at first, then more slowly as heat loss increases.
7. Check reality
Once boiling begins, the model changes.
Step 17: Why Differential Equations Matter
Differential equations are not just abstract formulas.
They describe how systems evolve.
In this example:
- heat enters,
- heat leaves,
- temperature changes,
- eventually boiling occurs.
The same structure appears in many other systems.
Health Example
A similar model can describe drug concentration in the bloodstream.
Let:
$$C(t)=\text{drug concentration}$$
If the body removes the drug at a rate proportional to concentration:
$$\frac{dC}{dt}=-kC$$
The solution is:
$$C(t)=C_0e^{-kt}$$
This is the same exponential structure that appears in heating and cooling.
Supply Chain Example
A similar model can describe inventory depletion.
Let:
$$I(t)=\text{inventory level}$$
If inventory leaves at a rate proportional to current stock:
$$\frac{dI}{dt}=-kI$$
Then:
$$I(t)=I_0e^{-kt}$$
If replenishment is added, the model becomes:
$$\frac{dI}{dt}=R-kI$$
This resembles the stove heating model:
$$\frac{dT}{dt}=\frac{P}{mc}-k(T-T_a)$$
The same mathematical structure appears in different fields.
Conclusion
Heating 200 ml of water on a stove may seem like a simple everyday action, but it contains the full logic of mathematical modeling.
We begin with a physical principle:
Energy entering minus energy leaving equals energy stored.
That principle leads to the differential equation:
$$mc\frac{dT}{dt}=P-h(T-T_a)$$
Solving it gives:
$$T(t)=T_\infty-(T_\infty-T_0)e^{-kt}$$
This solution tells us how the water temperature rises over time.
A simple model assumes constant power and gives linear heating.
A better model includes heat loss and gives exponential approach toward a limiting temperature.
Once boiling begins, the model changes again because added heat goes into phase change rather than temperature increase.
This is how applied mathematics works.
We start with reality.
We make assumptions.
We translate the situation into equations.
We solve the equations.
Then we interpret the solution carefully.
That is the power of differential equations.
References
Boyce, W. E., & DiPrima, R. C. Elementary Differential Equations and Boundary Value Problems.
Çengel, Y. A. Heat and Mass Transfer: Fundamentals and Applications.
Kreith, F., Manglik, R. M., & Bohn, M. S. Principles of Heat Transfer.
Newton, I. Philosophiae Naturalis Principia Mathematica.
Incropera, F. P., DeWitt, D. P., Bergman, T. L., & Lavine, A. S. Fundamentals of Heat and Mass Transfer.
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