riaology
Unit 3 · Lesson 2 of 6

A hot cup approaches the room temperature.

Cooling depends on the temperature gap. Change the initial temperature and rate constant to see a family of physical solutions.

Change the input. Follow the mathematics.

What you'll be able to do
  • Set up the model and calculate its temperature.
  • Connect the equation to the changing visual.
  • Explain the assumptions and identify a misleading interpretation.
Predict before exploring

What will change if you increase time? Predict the direction, then explain the response.

A model you can changeMove one control at a time.
Coffee cup at 42.1 degrees CelsiusSurroundings: 22°C
Time (min) versus Temperature (°C). Temperature, Ambient. Selected temperature: 42.07277 °C. Separating d(T−20)/(T−20)=−k dt gives exponential approach to equilibrium. Temperature gaps decay, not absolute Celsius temperature.0040158030Temperature (°C)Time (min)
TemperatureAmbient · dashed

temperatureT(10)=20+(80−20)e−1=42.0728 ∘CT(10)=20+(80-20)e^{-1}=42.0728\,{}^\circ\mathrm C

Separating d(T−20T-20)/(T−20T-20)=−k dt gives exponential approach to equilibrium. Temperature gaps decay, not absolute Celsius temperature.

Assumptions, units, and limits

Lumped uniform-temperature object, ambient 20°C, constant k; evaporation and heat generation neglected.

Slope field for cooling, with current initial-value trajectory. Segment angles account for axis scale.State (same units as main plot)Time · 0 to 30 min

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

01

From the situation to an equation

Separate the temperature-gap variable from time, integrate its reciprocal, and use the initial condition to fix the constant. Exponentiation gives the decaying gap. The ambient temperature is an equilibrium solution and the time constant is 1/k.

T′=−k(T−20),T(t)=20+(T0−20)e−ktT'=-k(T-20),\quad T(t)=20+(T_0-20)e^{-kt}
02

Keep the reasoning visible

ln⁡∣T−20∣=−kt+C,τ=1/k\ln|T-20|=-kt+C,\quad\tau=1/k

Separating d(T−20T-20)/(T−20T-20)=−k dt gives exponential approach to equilibrium. Temperature gaps decay, not absolute Celsius temperature.

03

What makes this model useful?

Lumped uniform-temperature object, ambient 20°C, constant k; evaporation and heat generation neglected.

The exponential decays the gap from ambient, not the absolute Celsius reading. This lumped model assumes the object is nearly uniform in temperature.

A worked example

Calculate the starting scenario

  1. Use time = 10 min; cooling constant k = 0.1 1/min; initial temperature = 80 °C.
  2. T(10)=20+(80−20)e−1=42.0728 ∘CT(10)=20+(80-20)e^{-1}=42.0728\,{}^\circ\mathrm C
  3. Separating d(T−20T-20)/(T−20T-20)=−k dt gives exponential approach to equilibrium. Temperature gaps decay, not absolute Celsius temperature. The checked quantity is temperature.
Make the idea your own

Calculate. Compare. Explain.

Numeric checks allow 1% rounding error. Values near zero use a small absolute tolerance. Work through the setup before opening a solution.

Calculate and explain

Rebuild the starting calculation

Calculate the temperature using time 10 min, cooling constant k 0.1 1/min, initial temperature 80 °C. State the meaning of its sign and units.

Checked quantity: temperature · °C. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Separate the temperature-gap variable from time, integrate its reciprocal, and use the initial condition to fix the constant. Exponentiation gives the decaying gap. The ambient temperature is an equilibrium solution and the time constant is 1/k.

Explore the visual solution
Coffee cup at 42.1 degrees CelsiusSurroundings: 22°C
Time (min) versus Temperature (°C). Temperature, Ambient. Selected temperature: 42.07277 °C. Separating d(T−20)/(T−20)=−k dt gives exponential approach to equilibrium. Temperature gaps decay, not absolute Celsius temperature.0040158030Temperature (°C)Time (min)
TemperatureAmbient · dashed

temperatureT(10)=20+(80−20)e−1=42.0728 ∘CT(10)=20+(80-20)e^{-1}=42.0728\,{}^\circ\mathrm C

Separating d(T−20T-20)/(T−20T-20)=−k dt gives exponential approach to equilibrium. Temperature gaps decay, not absolute Celsius temperature.

Assumptions, units, and limits

Lumped uniform-temperature object, ambient 20°C, constant k; evaporation and heat generation neglected.

Slope field for cooling, with current initial-value trajectory. Segment angles account for axis scale.State (same units as main plot)Time · 0 to 30 min

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Separate the temperature-gap variable from time, integrate its reciprocal, and use the initial condition to fix the constant. Exponentiation gives the decaying gap. The ambient temperature is an equilibrium solution and the time constant is 1/k.

T′=−k(T−20),T(t)=20+(T0−20)e−ktT'=-k(T-20),\quad T(t)=20+(T_0-20)e^{-kt}
Calculate and explain

Apply the idea to a new scenario

For T₀=60°C and k=0.05k=0.05 min⁻¹, find temperature after 20 min.

Checked quantity: temperature · °C. Follow the definition for finite-cutoff and numerical-model questions.

Hint

Separate the temperature-gap variable from time, integrate its reciprocal, and use the initial condition to fix the constant. Exponentiation gives the decaying gap. The ambient temperature is an equilibrium solution and the time constant is 1/k.

Explore the visual solution
Coffee cup at 34.7 degrees CelsiusSurroundings: 22°C
Time (min) versus Temperature (°C). Temperature, Ambient. Selected temperature: 34.71518 °C. Separating d(T−20)/(T−20)=−k dt gives exponential approach to equilibrium. Temperature gaps decay, not absolute Celsius temperature.0030156030Temperature (°C)Time (min)
TemperatureAmbient · dashed

temperatureT(20)=20+(60−20)e−1=34.7152 ∘CT(20)=20+(60-20)e^{-1}=34.7152\,{}^\circ\mathrm C

Separating d(T−20T-20)/(T−20T-20)=−k dt gives exponential approach to equilibrium. Temperature gaps decay, not absolute Celsius temperature.

Assumptions, units, and limits

Lumped uniform-temperature object, ambient 20°C, constant k; evaporation and heat generation neglected.

Slope field for cooling, with current initial-value trajectory. Segment angles account for axis scale.State (same units as main plot)Time · 0 to 30 min

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The controls explore variations. These written steps retain the original question's inputs.

Separate the temperature-gap variable from time, integrate its reciprocal, and use the initial condition to fix the constant. Exponentiation gives the decaying gap. The ambient temperature is an equilibrium solution and the time constant is 1/k.

T′=−k(T−20),T(t)=20+(T0−20)e−ktT'=-k(T-20),\quad T(t)=20+(T_0-20)e^{-kt}
Challenge the interpretation

Which explanation survives the model?

Choose the statement you can justify, then use the visual to explain why the others fail.

Test it with the visual
Coffee cup at 42.1 degrees CelsiusSurroundings: 22°C
Time (min) versus Temperature (°C). Temperature, Ambient. Selected temperature: 42.07277 °C. Separating d(T−20)/(T−20)=−k dt gives exponential approach to equilibrium. Temperature gaps decay, not absolute Celsius temperature.0040158030Temperature (°C)Time (min)
TemperatureAmbient · dashed

temperatureT(10)=20+(80−20)e−1=42.0728 ∘CT(10)=20+(80-20)e^{-1}=42.0728\,{}^\circ\mathrm C

Separating d(T−20T-20)/(T−20T-20)=−k dt gives exponential approach to equilibrium. Temperature gaps decay, not absolute Celsius temperature.

Assumptions, units, and limits

Lumped uniform-temperature object, ambient 20°C, constant k; evaporation and heat generation neglected.

Slope field for cooling, with current initial-value trajectory. Segment angles account for axis scale.State (same units as main plot)Time · 0 to 30 min

Each segment gives the local rate at that time and state. The red trajectory follows the chosen initial condition; Euler uses finite tangent steps.

The controls specify a new scenario. Curves are calculated from the equations; physical illustrations are schematic.

The reasoning. Separate the temperature-gap variable from time, integrate its reciprocal, and use the initial condition to fix the constant. Exponentiation gives the decaying gap. The ambient temperature is an equilibrium solution and the time constant is 1/k.

Your learning path

Can you explain the connection?

Mark complete when you can set up the calculation and explain what the visual is showing. Opening a page never completes it automatically.

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