Newton's Law of Cooling & Radiation Detectors

Cooling slows as a body nears room temperature; how to detect radiation.

8 minUpdated 22 October 2026Newton's law of coolingbolometerthermopile

Lesson at a glanceShow the key points

For small T − T₀: T⁴ − T₀⁴ ≈ 4T₀³(T − T₀); radiation and convection losses ∝ A(T − T₀).
Newton's law: dT/dt = −bA(T − T₀), or dθ/dt = −k(θ − θ₀) (28.9).
Bolometer: four platinum grids in a Wheatstone bridge; radiation heats two and unbalances it.
Thermopile: antimony–bismuth thermocouples in series (Seebeck effect), hot junctions blackened.

A cup of tea cools quickly at first, then lingers lukewarm. Newton found the simple rule behind this, and two instruments let us detect radiation directly.

From radiation to Newton’s law

The net radiation loss is eσA(T⁴ − T₀⁴). If T = T₀ + ΔT with ΔT small:

T^4 - T_0^4 = T_0^4\left[\left(1 + \frac{\Delta T}{T_0}\right)^4 - 1\right] \approx 4T_0^3\,\Delta T

So radiation loss ≈ b₁A(T − T₀). Convection loss for small differences is also proportional to A(T − T₀). With mass m and specific heat s, the rate of fall of temperature is

Newton's law of cooling(28.9)

\frac{dT}{dt} = -bA(T - T_0) \quad\text{or}\quad \frac{d\theta}{dt} = -k(\theta - \theta_0)

valid for small temperature differences; works equally in °C since only differences appear

Interactive

Newton's law of cooling

For small temperature differences, a body cools at a rate proportional to how much hotter it is than its surroundings: dθ/dt = −k(θ − θ₀). So it cools fast at first and more and more slowly, approaching room temperature. Each 10° step takes longer than the one before.

70°C → 60°C takes5.0 min

60°C → 50°C takes6.7 min

Cooling constant

k

0.050/min

Starting temperature80°C

Room temperature25°C

Worked example

Using average temperatures

Soup cools from 80°C to 70°C in 4.0 min in a room at 25°C. How long does it take to cool from 70°C to 60°C?

Detecting thermal radiation

The bolometer

Four identical grids cut from thin platinum foil form the arms of a Wheatstone bridge, balanced when no radiation falls. Grid 1 faces the radiation, arranged so whatever passes through its gaps lands on grid 4; grids 2 and 3 are shielded. Radiation heats grids 1 and 4, raising R₁ and R₄, so R₁R₄ > R₂R₃: the bridge unbalances and the galvanometer deflects, by an amount related to the radiation. Sealing it in an evacuated bulb makes it more sensitive.

The thermopile

Join two different metals at two junctions with a galvanometer in the circuit. If the junctions are at different temperatures, a current flows (the Seebeck effect): a thermocouple. A thermopile connects many antimony–bismuth thermocouples in series for sensitivity, with all the hot junctions on one blackened face exposed to the radiation (often behind a collecting cone) and the cold junctions shielded on the other.

What to remember

1For small differences, loss by radiation and convection ∝ A(T − T₀): dθ/dt = −k(θ − θ₀).
2Hot bodies cool fast at first, then slowly; use average temperatures over each interval in problems.
3Bolometer: radiation unbalances a bridge of platinum grids. Thermopile: many thermocouples in series, hot junctions blackened.

Check yourself

A body at 60°C in a 20°C room cools at 2 °C/min. At 40°C it cools at about…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadNewton's law of coolingHyperPhysics ReadNewton's law of coolingWikipedia ReadBolometerWikipedia ReadThermopileWikipedia

PreviousWien's & Stefan's Laws Next chapter Electric Field & Potential

Newton's Law of Cooling & Radiation Detectors

Cooling slows as a body nears room temperature; how to detect radiation.

8 minUpdated 22 October 2026Newton's law of coolingbolometerthermopile

Lesson at a glanceShow the key points

For small T − T₀: T⁴ − T₀⁴ ≈ 4T₀³(T − T₀); radiation and convection losses ∝ A(T − T₀).
Newton's law: dT/dt = −bA(T − T₀), or dθ/dt = −k(θ − θ₀) (28.9).
Bolometer: four platinum grids in a Wheatstone bridge; radiation heats two and unbalances it.
Thermopile: antimony–bismuth thermocouples in series (Seebeck effect), hot junctions blackened.

From radiation to Newton’s law

The net radiation loss is eσA(T⁴ − T₀⁴). If T = T₀ + ΔT with ΔT small:

T^4 - T_0^4 = T_0^4\left[\left(1 + \frac{\Delta T}{T_0}\right)^4 - 1\right] \approx 4T_0^3\,\Delta T

So radiation loss ≈ b₁A(T − T₀). Convection loss for small differences is also proportional to A(T − T₀). With mass m and specific heat s, the rate of fall of temperature is

Newton's law of cooling(28.9)

\frac{dT}{dt} = -bA(T - T_0) \quad\text{or}\quad \frac{d\theta}{dt} = -k(\theta - \theta_0)

valid for small temperature differences; works equally in °C since only differences appear

Interactive

Newton's law of cooling

70°C → 60°C takes5.0 min

60°C → 50°C takes6.7 min

Cooling constant

k

0.050/min

Starting temperature80°C

Room temperature25°C

Worked example

Using average temperatures

Soup cools from 80°C to 70°C in 4.0 min in a room at 25°C. How long does it take to cool from 70°C to 60°C?

Detecting thermal radiation

The bolometer

The thermopile

What to remember

1For small differences, loss by radiation and convection ∝ A(T − T₀): dθ/dt = −k(θ − θ₀).
2Hot bodies cool fast at first, then slowly; use average temperatures over each interval in problems.
3Bolometer: radiation unbalances a bridge of platinum grids. Thermopile: many thermocouples in series, hot junctions blackened.

Check yourself

A body at 60°C in a 20°C room cools at 2 °C/min. At 40°C it cools at about…

From radiation to Newton’s law#

Detecting thermal radiation#

The bolometer

The thermopile

Check yourself

Keep exploring

From radiation to Newton’s law#

Detecting thermal radiation#

The bolometer

The thermopile

Check yourself

Keep exploring

From radiation to Newton’s law

Detecting thermal radiation

From radiation to Newton’s law

Detecting thermal radiation