Newton’s law of cooling calculator

What is Newton’s law of cooling and its formula?

Newton’s law of cooling says that an object’s temperature changes at a rate proportional to the difference between its temperature and the temperature of its surroundings. Solving that gives T(t) = Ts + (T₀ − Ts)·e^(−kt): Ts is the surroundings, T₀ the starting temperature and k a positive cooling constant per unit of time. The same formula describes warming when T₀ is below Ts. OpenStax Calculus Volume 2 (section 4.3) writes the law as dT/dt = k(T − Ts) and uses a negative k in its examples (−0.08338, −0.125), so its k is ours with the opposite sign.

How do I find the cooling constant k?

Take one more reading after a known time: k = ln((T₀ − Ts) ÷ (T₁ − Ts)) ÷ t₁. OpenStax’s pizza example: a pizza leaves a 200°F oven in a 75°F kitchen and is 190°F one minute later, so k = ln(125 ÷ 115) = 0.08338 per minute (the book prints −0.08338). To cool to 150°F it needs t = ln(125 ÷ 75) ÷ k, which is 6.126 minutes without rounding k (the book prints 6.12). Use the “Cooling constant k” tab to find k, then the other tabs to use it.

How long does a hot drink take to cool down?

It depends on k, which depends on the cup, the lid, stirring and the air in the room, so measure it with two readings rather than guessing. For OpenStax’s coffee exercise (70°C coffee, 20°C room, k = 0.125 per minute) the law gives T = 20 + 50·e^(−0.125t): 34.3°C (93.8°F) after 10 minutes. The same coffee in °F is 158°F in a 68°F room: use the °F option and the answer is 93.8°F. Its gap to the room halves every ln 2 ÷ 0.125 = 5.5 minutes.

What are the half-way time and the time constant?

The gap between the object and its surroundings shrinks by the same fraction in equal times. It halves every ln 2 ÷ k (the half-way time, 69.3% of 1/k), and falls to 36.8% after one time constant 1/k, 13.5% after two, 5.0% after three and 0.7% after five. So the gap is down to 1% of its starting size after about 4.6 time constants, and it never quite closes: the law only approaches the surroundings and cannot cross them.

When is Newton’s law of cooling accurate?

It is a model, not an exact law. It assumes the heat-transfer coefficient stays constant, which is only approximately true when the temperature difference is small; for natural convection and for radiation the rate does not depend on the difference in such a simple way. It also treats the whole object as one temperature, which is reasonable for a stirred liquid or a small thin object (heat-transfer texts use a Biot number below about 0.1 as the rule of thumb), but not for a thick roast cooling from the outside in. Evaporation, boiling, freezing and a changing room temperature all break it. Treat k as a property of this object in these surroundings, not a constant of nature.

Can the calculator work with warming, as well as cooling?

Yes. When the start temperature is below the surroundings the object warms toward them by the same formula. An OpenStax exercise: ice cream at 10°F in an 80°F room rises 10°F in 10 minutes, which gives k = ln(70 ÷ 60) ÷ 10 = 0.01542 per minute and 40°F after 36.3 minutes. For how much heat is needed to change the temperature, use the specific heat calculator; for melting or boiling, the latent heat calculator; for heat flowing through a wall, the heat conduction calculator.

Sources: OpenStax Calculus, Volume 2, section 4.3, “Separable Equations” (Newton’s law of cooling, the pizza example and the exercises with the cake, the ice cream base and the coffee); University of Waterloo, ECE 309 lecture notes, chapter 5 (the lumped-capacitance model and its Biot number rule of thumb); Wikipedia, “Newton’s law of cooling” (limits of a constant heat-transfer coefficient). Related: specific heat calculator, latent heat calculator and heat conduction calculator.