Newton's Law of Cooling Calculator
Enter initial temperature, environment temperature, cooling constant and time to compute the temperature T(t)=T_env+(T₀-T_env)·e^(-kt).
Input Data
Results
At a glance:Newton's law of cooling (Newton, 1701) is the empirical relation dT/dt = −k·(T − T_env) for an object in a fluid, where T is the object temperature, T_env the environment temperature and k the cooling constant (1/s or 1/min) determined by surface properties, convection coefficient h, area A and heat capacity C via k = hA/C. With T(0) = T₀ the solution is T(t) = T_env + (T₀ − T_env)·e^(−k·t). As t→∞, T→T_env; at t = t_1/2 = ln(2)/k the temperature difference halves (half-life). It applies for forced convection or small temperature differences; for large temperature differences natural convection (k grows with ΔT) or radiation must be added.
Formula
Differential: dT/dt = −k·(T − T_env)
Integral: T(t) = T_env + (T₀ − T_env)·e^(−k·t)
Cooling constant: k = hA/C
Half-life: t_1/2 = ln(2)/k ≈ 0.693/k
Time constant: τ = 1/k
$$\frac{dT}{dt} = -k(T-T_{\text{env}}), \quad T(t) = T_{\text{env}} + (T_0 - T_{\text{env}})\,e^{-kt}, \quad k = \frac{hA}{C}, \quad t_{1/2} = \frac{\ln 2}{k}$$How to Use
- Enter the initial temperature T₀ (°C) and environment temperature T_env (°C).
- Enter the cooling constant k (1/min) and elapsed time t (min).
- The calculator shows the final temperature T(t), k (1/s), half-life t_1/2 and initial temperature difference.
Case Studies
Coffee cooling estimate
95°C coffee in a 20°C room, k ≈ 0.05/min.
To reach 60°C: t = ln(75/40)/0.05 ≈ 13.6 min.
To reach 40°C: t ≈ 26.7 min. A lid or thermos lowers k to 0.005–0.01, keeping it warm over an hour.
Forensic time of death
Body found at 30°C, room 20°C, normal 37°C, k ≈ 0.1/hr.
T(t) = 20 + 17·e^(−0.1t) = 30 → e^(−0.1t) = 10/17.
t ≈ 5.34 hr, i.e. death about 5h20m earlier; adjust for clothing, body type and humidity.
FAQ
When does Newton's law apply?
For forced convection or small-temperature-difference natural convection with a constant environment temperature and temperature-independent material properties. Large temperature differences (k grows with ΔT^(1/3)) or significant radiation require correction.
Why is the cooling curve exponential?
From dT/dt = −k(T−T_env), the larger the difference the faster the cooling, giving negative feedback so the difference decays as e^(−kt). Plot ln(T−T_env) vs t to get a straight line of slope −k, a standard way to measure k experimentally.
What determines k?
k = hA/C, where h is the convection coefficient (natural ~5–25, forced ~25–250 W/(m²·K)), A the surface area and C = mc the heat capacity. Larger area or stronger convection raises k; larger heat capacity lowers it.
How to measure k experimentally?
Record temperature vs time and plot ln(T − T_env) against t; the slope is −k. Keep T_env constant, ensure uniform internal temperature (Bi < 0.1), average multiple runs, and verify k is constant across differences.
How does radiation differ from Newton's law?
Newton's law models convection Q = hAΔT (proportional to ΔT, exponential decay). Radiation Q = εσA(T⁴ − T_env⁴) scales with the fourth power of absolute temperature and dominates above ~300°C. The total is dT/dt = −[hA(T−T_env) + εσA(T⁴−T_env⁴)]/C, no longer exponential, needing numerical integration.
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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.