Coffee Cup Cooling or Warming
A stock-flow model demonstrating Newton's law of cooling and warming, where the rate of temperature change is proportional to the discrepancy between the object and ambient temperature.
THE IDEA
Coffee Cup Cooling or Warming
This model illustrates Newton's law of cooling (and warming) using a simple stock-and-flow structure. The key variable is the temperature of the coffee in the cup, which changes over time based on the temperature difference between the coffee and the room.
Cooling
- Stock: coffee temperature(t) = coffee temperature(t - dt) - (cooling × dt)
- Initial values: 100°C, 80°C, and 60°C (three comparative runs)
- Time unit: minutes (dt = 1 minute)
- Run time: 8 minutes
- Outflow: cooling = discrepancy × 10%
- Converters:
- discrepancy = coffee temperature - room temperature
- room temperature = 18°C
Warming
- Stock: coffee temperature(t) = coffee temperature(t - dt) + (heating × dt)
- Initial values: 0°C, 5°C, and 10°C (three comparative runs)
- Inflow: heating = discrepancy × 10%
- Converters:
- discrepancy = room temperature - coffee temperature
- room temperature = 18°C
Both processes follow the same principle: the rate of change is directly proportional to the temperature gap. In cooling, the outflow reduces the stock; in warming, the inflow increases it. The coefficient of 10% per minute determines how quickly equilibrium is approached in this model.
When to use it
Predicting how long a hot beverage takes to cool to a drinkable temperature.
Designing a heating or cooling system where the rate depends on temperature difference.
Understanding how a smaller temperature difference slows warming or cooling in this model.
FROM THEORY TO EVERYDAY LIFE
In practice
Cooling Hot Coffee (Initial 100°C)
Starting at 100°C, the coffee loses heat quickly at first due to a large discrepancy (82°C), with cooling rate 8.2°C/min. As it approaches 18°C room temperature, the rate slows.
Warming Cold Coffee (Initial 0°C)
Starting at 0°C, the coffee gains heat rapidly (discrepancy 18°C, heating rate 1.8°C/min). After 8 minutes, it remains below 18°C, with warming slowing as the temperature gap shrinks.
A MOMENT TO REFLECT
Put it to work.
If you started with coffee at 90°C instead of 100°C, how would the cooling curve change?
How long would it take for the coffee to reach within 1°C of room temperature?
What physical factors might cause the 10% rate constant to be different in reality?
CONTINUE THE CONVERSATION
Source texts
- Thinking in Systems A Primer