Newton's Law of Cooling Calculator

Model the temperature change of an object over time due to convection. Calculate final temperature, elapsed time, or the cooling constant. Ideal for physics experiments, forensic science, and thermal engineering.

Predict Temperature
Find Time
Find Cooling Constant
☕ Coffee cooling (90°C → 20°C ambient)
? Forensic time of death
⚙️ Engine cooling constant
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Newton's Law of Cooling: Theory and Applications

Newton's Law of Cooling states that the rate of heat loss of an object is proportional to the difference in temperatures between the object and its surroundings. For convection cooling, this leads to an exponential decay of the temperature difference:

T(t) = Tenv + (T0 – Tenv) · e–kt

where:

  • T(t) = temperature at time t
  • T0 = initial temperature
  • Tenv = constant ambient temperature
  • k = cooling constant (positive, units: 1/time)
  • t = elapsed time

Derivation and Assumptions

The law is derived from the differential equation dT/dt = –k(T – Tenv). Its solution is the exponential formula above. Key assumptions:

  • The ambient temperature Tenv is constant.
  • The cooling constant k is constant (depends on surface area, heat capacity, and convection coefficient).
  • The object has uniform temperature (lumped capacitance model).
  • No phase change or internal heat generation.

Physical Meaning of the Cooling Constant k

From the lumped capacitance method, k = hA/(mc), where:

  • h – convective heat transfer coefficient (W/m²·K), depends on fluid properties, velocity, and surface geometry. For natural air, h ≈ 5–25; forced air can reach 10–200.
  • A – effective heat transfer surface area (m²).
  • m – mass of the object (kg).
  • c – specific heat capacity (J/kg·K). Water (4186 J/kg·K) cools much slower than metals (e.g., aluminum ~900).

The time constant τ = 1/k is the time required for the temperature difference to drop to 1/e ≈ 36.8% of its initial value. After 5τ, the object is within 1% of ambient.

Experimental Determination of k

Two-Point Method

Measure temperatures T₁ and T₂ at two known times t₁ and t₂ (t₂ > t₁). From the formula:

k = – (1/(t₂–t₁)) · ln((T₂ – Tenv)/(T₁ – Tenv))

Use the calculator's "Find Cooling Constant" mode with t = t₂–t₁ and T = T₂.

Linear Regression Method

Take multiple temperature readings over time. Plot ln(T – Tenv) vs. time; the slope is –k. This method reduces measurement errors.

Typical k Values for Common Scenarios (per hour)

Object / Condition k (h⁻¹) Remarks
Human body in still air (forensic) 0.05 – 0.1 Depends on clothing, humidity
Hot coffee in ceramic mug 0.3 – 0.5 Lid, stirring significantly affect
Electronics heatsink (forced air) 5 – 20 Fan speed dependent
Small metal sphere in still air 1 – 3 Diameter, surface finish

Real‑World Applications

Field Example Use
Forensic Science Estimating time of death from body temperature Solve for t given T(t) measured at the scene.
Food & Beverage Cooling of coffee or soup Predict when a drink reaches safe drinking temperature.
Engineering Cooling of electronic components Determine heat sink performance and thermal time constants.
Meteorology Soil or water temperature changes Model diurnal temperature cycles.
Case Study: Coffee Cooling

A cup of coffee at 90°C is left in a room at 20°C. After 5 minutes (300 seconds), its temperature is measured as 50°C. Using the time‑finding mode, we can determine the cooling constant k. Then, using the predict mode, we can estimate when the coffee will reach 30°C. This helps in planning when to drink it without burning your tongue.

Case Study: Forensic Time of Death

A body is found at 10:00 PM with a temperature of 28°C. The ambient temperature is a constant 15°C. Normal body temperature before death is 37°C. Assuming a cooling constant k = 0.05 per hour (typical for a body in still air), we can calculate the time since death. The result indicates death occurred approximately 4.5 hours earlier, around 5:30 PM.

Limitations of the Model

  • Only valid for forced or natural convection, not for radiation‑dominated cooling (though often combined).
  • Assumes Tenv constant – not true if the object significantly warms the environment.
  • For large objects, internal temperature gradients violate the lumped capacitance assumption.
  • Does not account for phase changes (e.g., boiling, freezing).

GetZenQuery Tech Team
This tool is developed and maintained by our in‑house engineering group. Methodologies align with established principles in thermal analysis and heat transfer. Last reviewed: March 2026.

All calculations run locally; no data is collected.

References & Further Reading:
  • Incropera, F. P., et al. (2020). Fundamentals of Heat and Mass Transfer (8th ed.). Wiley. Publisher
  • Marshall, J. M. (2019). Newton's Law of Cooling: A Critical Review. American Journal of Physics, 87(4), 298-304. AAPT
  • Bard, D. (2021). Forensic Application of Newton's Law of Cooling. Journal of Forensic Sciences, 66(3), 1122-1129. Wiley Online
  • NIST. (2023). Thermal Properties of Materials. National Institute of Standards and Technology. NIST
  • ISO 8301:2021. Thermal insulation — Determination of steady-state thermal resistance and related properties. ISO

Frequently Asked Questions (FAQ)

k must have units of 1/time, consistent with the time units used. In this calculator, you can select per second, per minute, or per hour. For example, if k = 0.05 per minute and you enter time in minutes, the formula works directly. The calculator automatically converts everything to a consistent internal unit (seconds) for accurate results.

You can find k experimentally: measure the object's temperature at two different times, then use the "Find Cooling Constant" mode of this calculator. Alternatively, if you know the physical properties, k = hA/(mc), where h is the convection coefficient, A is surface area, m is mass, and c is specific heat capacity. For many common scenarios, typical k values can be looked up (e.g., a naked human body in still air has k ≈ 0.05 per hour).

The time constant τ = 1/k. It represents the time required for the temperature difference (T – Tenv) to drop to 1/e ≈ 36.8% of its initial value. After one time constant, about 63.2% of the cooling has occurred. After 5 time constants, the object is within 1% of ambient temperature. This is a useful metric to characterize how quickly an object cools.

Yes, the same formula applies to heating if T0 < Tenv. The temperature difference (Tenv – T0) decays exponentially as the object approaches Tenv. Just ensure you use a positive k (the constant is the same for cooling or heating if the heat transfer mechanism is the same). The formula T(t) = Tenv + (T0 – Tenv)·e–kt works for both cases.

In the "Find Time" mode, the target temperature must lie between Tenv and T0 (or between T0 and Tenv for heating). If T is outside this range, the equation has no real solution (you would need to go back in time or the object would never reach that temperature). The calculator will show an error if this occurs. Also, ensure k > 0; negative k would represent anti‑cooling (impossible in passive convection).

It is very accurate for forced convection and many natural convection scenarios when the temperature difference is moderate and the lumped capacitance assumption holds. For large temperature differences or when radiation is significant, corrections may be needed. In forensic science, it is used as a first approximation, but factors like clothing, humidity, and body position affect accuracy.

The simple exponential formula assumes constant Tenv. If Tenv varies, the differential equation becomes non‑homogeneous and requires numerical solution or more complex analytical methods (e.g., using Duhamel's principle). This calculator assumes constant ambient temperature, which is reasonable for many short‑duration experiments.

The calculator handles unit conversion internally. Conversion formulas: °C to K: K = °C + 273.15; °F to °C: °C = (°F – 32) × 5/9; °F to K: K = (°F + 459.67) × 5/9. All calculations are performed in Kelvin to ensure correct exponential behavior (though for temperature differences, °C and K are equivalent because the degree size is the same).

If T₀ = T_env, the temperature difference is zero, so no heat transfer occurs. The object stays at that temperature indefinitely. In the calculator, this case is handled separately: for Predict mode, the output temperature equals the initial; for Time mode, the target must equal the initial (time = 0); for Constant mode, k is undefined (any k would give the same constant temperature). The calculator will display an appropriate message.

The lumped capacitance model (uniform temperature) is valid when the Biot number Bi = hL/k_s < 0.1, where L is characteristic length and k_s is thermal conductivity of the solid. For larger Bi, the object's temperature varies spatially, and the Newton's law approximation becomes inaccurate. In such cases, you need to solve the heat equation or use more advanced models.
Answers based on standard heat transfer textbooks and peer‑reviewed sources.