Newton's Cooling Rate Calculator
Enter initial and environment temperatures, cooling constant and time to compute the cooling rate dT/dt at any moment.
Input Data
Results
At a glance:Newton's law of cooling models the rate of temperature change of an object in a fluid: dT/dt = −k·(T − T_env), where T is the object's temperature, T_env the (constant) environment temperature, and k the cooling constant (1/s). Solving with T(0)=T₀ gives T(t) = T_env + (T₀ − T_env)·e^(−k·t). The cooling rate dT/dt equals −k·(T(t) − T_env), largest when the temperature difference is largest and decaying to zero as T→T_env. k depends on convection, surface area and heat capacity (k = hA/C).
Formula
Differential: dT/dt = −k·(T − T_env)
Solution: T(t) = T_env + (T₀ − T_env)·e^(−k·t)
Cooling rate: dT/dt = −k·(T(t) − T_env)
$$\frac{d\Delta T}{dt} = -k\Delta T, \quad T(t) = T_{\text{env}} + (T_0 - T_{\text{env}})e^{-kt}, \quad \Delta T(t) = \Delta T_0 e^{-kt}, \quad \tau = \frac{1}{k}, \quad t_{1/2} = \frac{\ln 2}{k}$$How to Use
- Enter the initial temperature T₀ and environment temperature T_env (°C).
- Enter the cooling constant k (1/s) and time t (s).
- The calculator returns T(t) and the instantaneous cooling rate dT/dt.
Case Studies
Cooling cup of coffee
T₀ = 95°C, T_env = 20°C, k = 0.0008/s.
At t = 0: dT/dt = −0.0008×75 ≈ −0.06 °C/s.
After 600 s: T ≈ 20 + 75·e^(−0.48) ≈ 66°C.
FAQ
When does Newton's law apply?
For forced convection or small temperature differences where the convection coefficient is roughly constant, and when the environment temperature is steady and large compared with the object.
Why does the cooling rate decay over time?
The rate is proportional to the temperature difference T − T_env, which shrinks exponentially, so dT/dt also decays exponentially toward zero.
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References
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.