What is Newton's law of cooling equation?
The equation is T(t) = Ts + (T0 minus Ts) times e to the power of minus k times t. Ts is the ambient (surrounding) temperature, T0 is the starting temperature of the object at time zero, k is the positive cooling coefficient, t is elapsed time, and T(t) is the temperature after time t. It says the temperature gap between the object and its surroundings shrinks exponentially, never quite reaching ambient.
How do I find the cooling coefficient k from two temperature measurements?
Measure the starting temperature T0, the ambient temperature Ts, then a later reading T1 taken at time t1. Rearranging the law gives k = minus ln((T1 minus Ts) divided by (T0 minus Ts)) divided by t1, using the natural logarithm. The fraction must be strictly between 0 and 1, which simply means the object has moved partway toward ambient without crossing it. Example: Ts = 20 C, T0 = 90 C, and T1 = 62.46 C after 10 minutes gives k = 0.05 per minute.
How long until my hot coffee reaches drinking temperature?
Use the time-to-target mode with your numbers. For example, coffee starting at 90 C in a 20 C room with a cooling coefficient of 0.05 per minute reaches 60 C after about 11.2 minutes: t = minus ln((60 minus 20) divided by (90 minus 20)) divided by 0.05 = minus ln(0.5714) divided by 0.05. This calculator solves that for any temperatures, times, and units.
What are the units of the cooling coefficient k?
k is a rate, so its units are per unit of time: per second, per minute, or per hour. A value of 0.05 per minute is the same as 3.0 per hour or about 0.00083 per second, and this calculator converts between them automatically. Smaller k means slower cooling: a well-insulated thermos has a tiny k, while a thin metal cup in a breeze has a large one.
Can Newton's law of cooling be used for warming objects too?
Yes. The same equation handles warming: if the object starts colder than the surroundings (T0 below Ts), the negative gap still shrinks exponentially and the object warms toward ambient. A cold baby bottle left on the counter at 25 C warming from 4 C with k = 0.03 per minute follows exactly the same math as cooling coffee.
What assumptions does Newton's law of cooling make?
Four main ones: the ambient temperature stays constant during the experiment, the object has a uniform internal temperature (no hot core or cold spots, valid for small or conductive objects and stirred liquids), heat loss is dominated by convection (the linear approximation of radiation), and k stays constant (no change in airflow, surface area, or phase). Break one of these and measured k will drift.
Why does my measured k change between experiments?
Because k is not a property of the liquid alone: it lumps together everything that moves heat, including cup material and thickness, exposed surface area, air drafts, evaporation, and stirring. Change the cup or open a window and you measure a different k. For repeatable results keep the container, volume, and airflow identical and fit k from your own two measurements rather than reusing a published number.
What are the time constant and the half-life of cooling?
The time constant tau = 1 divided by k is the time for the temperature gap to shrink to about 37 percent (1 over e) of its start. The half-life = ln(2) divided by k, about 0.693 times tau, is the time for the gap to halve: with k = 0.05 per minute the half-life is about 13.9 minutes, so a 70-degree gap becomes 35 degrees after 13.9 minutes and about 17.5 degrees after 27.7 minutes.
Does Newton's law of cooling include heat radiation?
Not exactly. The law assumes heat loss is proportional to the temperature difference, which is the convective regime. True thermal radiation follows the Stefan-Boltzmann law and scales with the fourth power of absolute temperature, so it cools faster than the linear model at large temperature gaps. For everyday gaps of a few dozen degrees around room temperature the linear approximation is usually good enough, and any small radiation contribution just gets absorbed into the fitted k.