Newton's Law of Cooling: Which Form to Use in Numericals

Newton's law of cooling says the rate at which a body loses heat is proportional to the difference between its temperature and the temperature of its surroundings. Newton published the idea in 1701, in "Scala graduum Caloris" (Wikipedia: Newton's law of cooling).

In symbols, with T the body's temperature, Tₛ the surroundings' temperature and k a positive constant:

dT/dt = −k(T − Tₛ)

We call T − Tₛ the temperature excess. It is the only quantity that matters in this law. A cup at 90 °C in a 25 °C room has an excess of 65 °C, and the cup cools fast. A cup at 30 °C in the same room has an excess of 5 °C, and it barely cools.

Solving the equation (separate the variables and integrate) gives the exponential form:

T = Tₛ + (T₀ − Tₛ)e^(−kt)

Here T₀ is the starting temperature. The constant k is 1/τ, where τ is the time constant of the body. In the lumped model τ = C/(hA), where C is the body's heat capacity, h the heat-transfer coefficient and A the surface area (Wikipedia). A body with a large surface and small heat capacity has a large k and cools quickly.

MHT-CET (the Maharashtra Common Entrance Test) publishes its physics syllabus in the "Syllabus and Marking Scheme" download on the CET Cell site.

Use the form that matches what the question asks and how far the temperature falls. This table is the decision we use before writing a single number.

To apply the test, compute the fall fraction:

f = (T_start − T_end) / (T_start − Tₛ)

If f is 0.3 or less, use the shortcut. If it is above 0.3, use the exponential.

The average-temperature shortcut treats the cooling rate as constant over the interval, equal to the rate at the mean of the start and end temperatures. For a body going from T₁ to T₂ in time t:

(T₁ − T₂)/t = k × [(T₁ + T₂)/2 − Tₛ]

A body cools from 80 °C to 64 °C in 5 minutes in a room at 20 °C. The time it takes to go from 64 °C to 52 °C is 5.13 minutes by either method. Here is the working.

Exponential method.

1. Excess at the start is 80 − 20 = 60 °C, and after 5 minutes it is 64 − 20 = 44 °C.
2. k = ln(60/44) / 5 = 0.3102 / 5 = 0.0620 per minute.
3. Excess at 52 °C is 32 °C. Time from 64 °C to 52 °C = ln(44/32) / k = 0.3185 / 0.0620 = 5.13 minutes.

Shortcut method.

1. First interval: the mean temperature is 72 °C, so the mean excess is 52 °C. The rate is 16/5 = 3.2 °C per minute. So k = 3.2 / 52 = 0.0615 per minute.
2. Second interval: the mean of 64 and 52 is 58 °C, so the mean excess is 38 °C. The rate is 0.0615 × 38 = 2.34 °C per minute.
3. Time = 12 / 2.34 = 5.13 minutes.

The two answers agree because both intervals have a similar fall fraction. The first has f = 16/60 = 0.27, and the second has f = 12/44 = 0.27. The errors in the two steps point the same way and mostly cancel. That cancellation is our inference from this example. It will not hold when the two steps have very different fall fractions.

As a sanity check on the exponential result, the excess halves every ln 2 / k = 0.693 / 0.0620 = 11.2 minutes. This figure is ours, worked from the example's k.

The shortcut stays within about 1% while the excess falls by 30% or less in one interval, then drifts to about 4% at 50% and about 11% at 70%. This table is our own calculation.

Here is how we worked it out. Suppose the excess falls by a fraction f in time t.

• The exact law gives kt = −ln(1 − f).
• The shortcut gives f = k × (1 − f/2) × t, so kt = f / (1 − f/2).
• The percentage error is the shortcut value minus the exact value, divided by the exact value.

The shortcut always underestimates kt. It replaces a curve with a straight line, and the curve bends upwards.

Is 1% fine for an exam? That depends on how far apart the options in a multiple-choice question sit. Our recommendation is to use the shortcut only up to f = 0.3. For anything bigger, the exponential costs you one extra logarithm and removes the doubt. Check the example above: the first interval had f = 0.27, which is within that limit.

Most lost marks come from using the wrong quantity, not from the calculus. Run these four checks before you submit. Each has a clear pass and fail.

1. Did you subtract the room temperature before taking any ratio? Pass: every number in your ratio is T − Tₛ. Fail: you wrote ln(80/64) instead of ln(60/44).
2. Did you average the temperatures, then subtract Tₛ? Pass: mean of 64 and 52 is 58, then 58 − 20 = 38. Fail: you subtracted Tₛ from only one of the two temperatures, or used the mean temperature (58 °C) directly as the excess instead of 58 − 20 = 38 °C.
3. Is the fall fraction f at most 0.3 when you use the shortcut? Pass: you computed f = (T_start − T_end)/(T_start − Tₛ) and it is 0.3 or less. Fail: you did not check.
4. Did the surroundings stay fixed? Pass: the question gives one room temperature for the whole problem. Fail: a fan is switched on, or the room warms up, partway through.

One more habit saves time. Temperature differences are the same in degrees Celsius and in kelvin, because the two scales have the same step size. So you can stay in °C throughout, and you do not need to convert.

For a general routine you can apply to any numerical, see our guide on how to solve physics numericals in four steps for HSC and MHT-CET.

Newton's law of cooling holds only when the surroundings stay at a constant temperature and the temperature difference is small enough. For heat lost by radiation, it holds only for very small temperature differences (Wikipedia).

Three limits are worth knowing:

• Radiation. Radiated heat depends on the fourth power of absolute temperature, not on the plain difference. A very hot body cools faster at first than the law predicts.
• Natural convection with large differences. The law is only approximately true. Dulong and Petit modified it in 1817 by adding an exponent for larger differences (Wikipedia).
• A body whose inside is not at one temperature. The simple model assumes uniform temperature inside the body. A Biot number below 0.1 typically means less than 5% error from the lumped model (Wikipedia). The Biot number compares how easily heat leaves the surface with how easily it moves inside the body.

The law also never lets the body reach the surrounding temperature in finite time. The term e^(−kt) gets smaller and smaller but never equals zero. In practice, the difference becomes too small to measure.

Practise five numericals on this law this week. In each, compute f first and then choose your method from the table above. That one habit fixes most errors on this topic.

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Choosing a class for 11th is a bigger decision than one chapter. Our guide to 11th science classes in Maharashtra compares costs and fit. If you want another heat-related chapter for practice, try our page on the kinetic theory of gases.