Solve T(t) = Ts + (T0 − Ts)·e−kt for the temperature of a cooling (or warming) object at any time, the time it takes to reach a target temperature, or the cooling constant from two timed readings — plus the cooling curve.
Inputs
The steady surrounding temperature the object cools (or warms) toward.
How quickly the gap to ambient closes — bigger k means faster cooling. Find it on the Cooling Constant tab if you only have two thermometer readings.
Presets (illustrative values)
Result
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Enter the inputs to compute a result.
Difference from ambient—
Toward equilibrium—
Cooling half-life—
Time constant (τ)—
Inputs
Must be strictly between the ambient and starting temperatures.
Time to reach target
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Enter the inputs to compute a result.
Cooling half-life—
Time constant (τ)—
Two timed readings
No need to know the exact temperature at t = 0 — just two readings taken while it cools.
Cooling constant
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Enter the inputs to compute a result.
Cooling half-life—
Time constant (τ)—
Model assumptions
Ambient temperature is constant.
The object is treated as having a uniform internal temperature.
The cooling constant remains constant over the modeled interval.
Evaporation, radiation changes, phase changes, and active heating are not modeled.
Results are estimates and may differ from real measurements because of thermometer calibration, sensor response time, evaporation, container geometry, and changing airflow.
Newton's law of cooling describes one of the most common physical processes around us: how a hot mug of coffee, a fresh-baked loaf, or a warm car engine loses heat to its surroundings.
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Walk-through
How to Use This Calculator
3 steps▸
1
Pick a tab for what you know
Use Temperature when you know the ambient temperature, the starting temperature, and the cooling constant, and want the temperature at a given time. Use Time to Target when you want to know how long it takes to reach a specific temperature. Use Cooling Constant when you have two timed thermometer readings and want to find k.
2
Enter the values and matching units
Type the ambient temperature, the relevant time or temperature values, and pick a temperature unit (°F or °C) and a time unit (seconds, minutes, or hours) — the cooling constant k is assumed to share whichever time unit you select. Tap a preset like Coffee or Bath water to fill in illustrative values instantly.
3
Read the result, stats, and curve
The result card shows the solved value plus the temperature gap, percent toward equilibrium, cooling half-life, and time constant. On the Temperature tab, the chart plots the full cooling curve — how the object's temperature approaches ambient over time.
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Reference
Formula & Methodology
4 formulas▸
Newton's law of cooling
T(t) = Ts + (T0 − Ts) · e^(−kt)
T(t) is the object's temperature at time t, Ts is the ambient (surrounding) temperature, T0 is the starting temperature, and k is the cooling constant. The temperature gap (T0 − Ts) shrinks exponentially — it never technically reaches zero, but gets arbitrarily close.
Solve for elapsed time
t = −ln((T − Ts) / (T0 − Ts)) / k
Rearranges the cooling law for time. The target T must sit strictly between Ts and T0 — the object passes through every temperature in that range exactly once on its way to equilibrium.
Solve for the cooling constant from two readings
k = ln((T1 − Ts) / (T2 − Ts)) / (t2 − t1)
Finds k from two timed temperature readings (t1, T1) and (t2, T2) without needing to know the exact temperature at t = 0 — useful when you start timing after the object is already cooling.
Cooling half-life and time constant
t½ = ln(2) / k · τ = 1 / k
The half-life t½ is how long it takes the temperature gap to close by half; the time constant τ is how long it takes the gap to close to about 36.8% (1/e) of its starting size. Both are the same relations used for radioactive decay and RC circuit discharge — Newton's law of cooling belongs to the same exponential-relaxation family.
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Glossary
Key Terms Explained
7 terms▸
Newton's law of cooling ↗The empirical rule that an object's rate of heat loss (or gain) is proportional to the difference between its own temperature and the ambient temperature. It produces the exponential relation T(t) = Ts + (T0 − Ts)e^(−kt) and holds well for convection-dominated cooling with a modest temperature difference.
Ambient temperature (Ts) ↗The steady temperature of the surrounding environment that the object cools or warms toward — the room, the air, the water bath. The model assumes this stays constant throughout the process.
Cooling constant (k) ↗A positive rate constant (units of inverse time) that sets how fast the temperature gap closes. It depends on the object's surface area, material, and the surrounding airflow — larger k means faster cooling. It is not a universal constant; it must be measured or fit for each object and setting.
Exponential decay ↗A process where the remaining quantity — here, the temperature gap above or below ambient — shrinks by the same fraction in each equal time interval. Newton's law of cooling, radioactive decay, and capacitor discharge all follow this same mathematical pattern.
Thermal equilibrium ↗The state where an object's temperature equals the ambient temperature and net heat flow stops. The exponential model approaches this asymptotically — it predicts the gap gets arbitrarily small but never mathematically reaches exactly zero.
Cooling curve ↗The graph of an object's temperature versus time as it cools (or warms) toward ambient — a curve that starts steep and flattens out as the temperature gap shrinks, characteristic of exponential decay.
Half-life (of cooling) ↗The time it takes for the temperature gap between the object and ambient to shrink by half, equal to ln(2)/k. Like radioactive half-life, it stays constant throughout the process regardless of how large the gap currently is.
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Scenarios
Real-World Examples
3 worked examples▸
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Coffee drinker timing a mug
Coffee cooling in a 70°F room
Ambient temperature 70°FStarting temperature 180°FCooling constant 0.05 /minElapsed time 10 min
After 10 minutes, the coffee has cooled to about 136.7°F — still 66.7°F above room temperature, roughly 39% of the way to equilibrium. The temperature gap halves roughly every 13.9 minutes (ln(2)/0.05), so the coffee cools quickly at first and then levels off.
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Someone checking when coffee is drinkable
Time until the coffee drops to a comfortable 140°F
Ambient temperature 70°FStarting temperature 180°FCooling constant 0.05 /minTarget temperature 140°F
Solving t = −ln((140−70)/(180−70))/0.05 gives about 9.0 minutes. Because the model is exponential, waiting a little longer to reach, say, 100°F takes much more time (about 23.1 minutes) — most of the cooling happens early, and the last stretch toward ambient is slow.
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Student measuring an unknown cooling constant
Finding k from two thermometer readings
Ambient temperature 70°FReading 1 155.7°F at 5 minReading 2 122.0°F at 15 min
k = ln((155.7−70)/(122.0−70)) / (15−5) ≈ 0.05 per minute — recovered without ever needing to know the exact starting temperature at t = 0. That k can now be plugged into the Temperature or Time to Target tab to project the object's temperature at any other time.
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Reference
Cite This Calculator
APA & MLA▸
Use either format to cite this calculator in a paper, report, or resource list.
Newton's law of cooling describes one of the most common physical processes around us: how a hot mug of coffee, a fresh-baked loaf, or a warm car engine loses heat to its surroundings. It belongs to the same family of exponential-relaxation math as radioactive decay and capacitor discharge — a quantity approaches a steady value at a rate proportional to how far it still has to go. This guide walks through the physics behind the formula, how to find the cooling constant experimentally, where the model is used in practice, and where its simplifying assumptions break down.
Why cooling is exponential, not linear
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It's tempting to think an object cools at a constant rate, but that isn't what actually happens. Newton's law of cooling states that the rate of heat loss is proportional to the current temperature difference from ambient — a hot object loses heat fast, and as it approaches room temperature, it loses heat more and more slowly. Solving that rate relationship gives the exponential formula T(t) = Ts + (T0 − Ts)e^(−kt): the temperature gap shrinks by the same fraction in every equal time interval, not by the same number of degrees. That's why a cup of coffee cools noticeably in its first five minutes but takes much longer to lose its last few degrees above room temperature — the closer it gets to equilibrium, the slower the remaining cooling becomes.
Finding the cooling constant experimentally
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The cooling constant k isn't a fixed physical constant like the speed of light — it depends on the object's shape, surface area, material, and the airflow around it, so it has to be measured for each situation. The Cooling Constant tab does this the practical way: take two temperature readings at two different times while the object cools, and solve k = ln((T1 − Ts)/(T2 − Ts)) / (t2 − t1). This works even if you started your stopwatch after the object was already cooling, because it only needs the relationship between two later readings, not the exact starting temperature. Once you have k, you can predict the temperature at any future time or work out how long until it reaches a target — that's the same math a forensic investigator uses (very approximately) to estimate how long a body has been cooling, and the same reasoning a home baker uses to guess how much longer a loaf needs before it's safe to slice.
Half-life, time constant, and the language of decay
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Because Newton's law of cooling is mathematically identical to radioactive decay, it inherits the same descriptive vocabulary. The half-life of cooling t½ = ln(2)/k is the time for the temperature gap to close by half — a coffee with k = 0.05 per minute has a cooling half-life of about 13.9 minutes, meaning it takes about 13.9 minutes to go from 66.7°F above room temperature to 33.3°F above, then another 13.9 minutes to reach 16.7°F above, and so on. The time constant τ = 1/k is the time for the gap to shrink to about 36.8% (1/e) of its start. Both numbers describe the same underlying process from different angles, and both stay constant throughout the cooling process regardless of how large the current gap is — a property unique to exponential decay.
Limits and edge cases
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Newton's law of cooling is a good approximation, not an exact law, and it rests on a few assumptions that can break down. It assumes the ambient temperature stays constant — a coffee cooling on a windy porch or in a car with the AC cycling on and off won't follow a clean exponential curve. It assumes the object's own temperature is uniform throughout (a thin cup of coffee qualifies; a thick roast with a hot center and cool crust does not, at least not early on). It also assumes heat loss is dominated by convection with a modest temperature difference — for very large temperature gaps, radiative heat loss (which scales with the fourth power of absolute temperature, per the Stefan-Boltzmann law) becomes significant and the simple exponential model under- or overestimates the cooling rate. The calculator also requires k > 0 and a target temperature strictly between the ambient and starting values on the Time to Target tab — an object can't reach a temperature it never passes through.
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Questions
Frequently Asked Questions
6 questions▸
What is the formula for Newton's law of cooling?+
T(t) = Ts + (T0 − Ts)·e^(−kt), where Ts is the ambient temperature, T0 is the starting temperature, k is the cooling constant, and t is elapsed time. It says the temperature gap between the object and its surroundings shrinks exponentially over time.
How do I find the cooling constant k?+
Take two temperature readings at two different times while the object cools and use the Cooling Constant tab: k = ln((T1 − Ts)/(T2 − Ts)) / (t2 − t1). You don't need to know the exact starting temperature at t = 0 — any two later readings work, as long as they're on the same side of the ambient temperature.
How do I find the time it takes to cool to a certain temperature?+
Use the Time to Target tab, which solves t = −ln((T − Ts)/(T0 − Ts)) / k for the elapsed time. The target temperature must sit strictly between the ambient and starting temperatures — the object passes through every value in that range exactly once.
What does thermal equilibrium mean here?+
It's the point where the object's temperature equals the ambient temperature and heat stops flowing. The exponential model predicts the gap gets arbitrarily small over time but, mathematically, never reaches exactly zero — in practice an object is considered 'at ambient' once the remaining gap is too small to matter.
What units does this calculator use?+
Temperatures can be entered in °F or °C — pick whichever unit you're working in on each tab. Time can be in seconds, minutes, or hours; the cooling constant k is assumed to share whatever time unit you select, since k's units are always the inverse of a time unit.
Does Newton's law of cooling work for heating too?+
Yes — the same formula applies whether the object starts below or above ambient. If T0 is less than Ts (for example, a cold drink warming up in a room), the math is identical; the object's temperature rises exponentially toward ambient instead of falling toward it.
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