Audited 29 Jul 2026·Last updated 28 Oct 2026·8 citations·Tier 2·0 uses

Black Hole Temperature Calculator (Hawking Radiation)

Find a black hole's Hawking temperature from its mass, or the mass from a temperature, with horizon radius, entropy, radiated power and evaporation time.

Black Hole Temperature Calculator

What are you starting from?
SI kilograms. Reference masses: Sun 1.98841 × 10³⁰ (PDG 2024) · Sagittarius A* 8.5442 × 10³⁶ · M87* 1.2925 × 10⁴⁰ · a hole at exactly the CMB temperature 4.5016 × 10²² · a hole evaporating in one age of the universe 1.7294 × 10¹¹. Must be greater than zero.
Used in the temperature → mass route. Thermodynamic (absolute) temperature only. A one-solar-mass hole sits at 6.17026 × 10⁻⁸ K; the cosmic microwave background is at 2.72548 K. Must be greater than zero.
Hawking temperature (K)
6.170 × 10⁻⁸
Hawking temperature in kelvin, T_H = ħc³/(8πGMk_B), for a non-rotating, uncharged black hole.
Mass (kg)
1.988 × 10³⁰
Mass (M☉)
1
Event-horizon radius (m)
2,953.2503
Entropy S / k_B (dimensionless)
1.049 × 10⁷⁷
Entropy S (J/K)
1.448 × 10⁵⁴
Radiated power — idealised (W)
9.008 × 10⁻²⁹
Evaporation time — idealised (years)
2.095 × 10⁶⁷
Evaporation time — idealised (s)
6.613 × 10⁷⁴
T_H ÷ T_CMB
2.264 × 10⁻⁸
Conventions and limits for this result
Schwarzschild (zero spin, zero charge) and semiclassical: the spacetime is treated classically while the radiation field is quantum. T_H and the Bekenstein–Hawking entropy are exact in that regime; a spinning Kerr hole of the same mass is colder. The radiated power and evaporation time are an IDEALISATION — a perfect blackbody of area 4πr_s² emitting photons only. Page (1976) computed the real emission for a hole hot enough to evaporate and found roughly 9.7 times more power once greybody factors and other particle species are included, so a true lifetime is about an order of magnitude shorter than the figure shown. Absorption of the cosmic microwave background is not modelled anywhere on this page. WARNING: this hole is colder than the 2.72548 K cosmic microwave background, so today it absorbs more energy than it radiates and is GROWING, not evaporating. Every black hole heavier than about 4.50 × 10²² kg — which is every black hole ever observed — is in this state. The evaporation time above is therefore not a prediction; it is what the lifetime would be in an empty, cold universe.

Background.

Compress the Sun into a sphere 5.9 kilometres across and it becomes a black hole with a Hawking temperature of 6.17 × 10⁻⁸ kelvin, sixty-two billionths of a degree above absolute zero. Shrink the mass to that of the Moon and the temperature climbs only to 1.67 K. Both are colder than the 2.72548 K microwave background that fills the sky, so both would take in more energy from their surroundings than they give out.

That inverse relationship is the core of the calculation. Stephen Hawking's 1975 paper showed that a horizon radiates like a warm body at a temperature fixed by its surface gravity, and for a hole with no spin and no charge this works out to T_H = ħc³/(8πGMk_B). Double the mass and the temperature halves; nothing about what fell in, or when, enters at all. Start from a mass to get the temperature, or from a temperature to get the mass that would produce it. Either way you also get the horizon radius, the Bekenstein–Hawking entropy, the radiated power and an evaporation time.

Those outputs are not equally solid. Temperature and entropy are exact results of the semiclassical theory for a non-rotating hole. Power and lifetime rest on a textbook simplification: the horizon treated as a perfect blackbody that emits only photons. Don Page's 1976 calculation of the real emission found about 9.7 times more power once neutrinos and gravitons are counted and the filtering effect of the hole's own gravitational field is included, so an evaporating hole really dies roughly ten times sooner than the printed lifetime. Neither figure allows for the microwave background feeding the hole, and a spinning hole of the same mass is colder than the temperature shown.

What is black hole temperature calculator?

Hawking temperature is the temperature of the thermal radiation that a distant observer would see coming from a black hole's horizon. It is set by the horizon's surface gravity, κ = c⁴/(4GM), through T_H = ħκ/(2πck_B), so for a non-rotating, uncharged hole the mass alone determines it. The companion quantity is the Bekenstein–Hawking entropy, which counts the horizon's area in units of four Planck areas and grows with the square of the mass.

How to use this calculator.

  1. Choose whether you are starting from a black hole's mass or from a Hawking temperature.
  2. For mass, enter kilograms. The field hint gives the Sun (1.98841 × 10³⁰ kg), Sagittarius A* and M87*; to convert solar masses, multiply by 1.98841 × 10³⁰.
  3. For temperature, enter kelvin on the absolute scale. Entering 2.72548 K finds the hole that is exactly as warm as the microwave background.
  4. Check T_H ÷ T_CMB before reading the lifetime. Below 1, the hole is gaining mass from the background today and the lifetime describes an empty, cold universe instead.
  5. Read the power and evaporation time alongside the note beneath them, which states the photon-only simplification and flags any hole short-lived enough to have evaporated already.

The formula.

T_H = ħc³ ⁄ (8πGMk_B) · S = 4πk_B GM² ⁄ (ħc) = k_B A c³ ⁄ (4ħG) · P = ħc⁶ ⁄ (15360πG²M²) · t = 5120πG²M³ ⁄ (ħc⁴)

Hawking's result is cleanest in natural units, where Wald's review states it as T = κ/2π and the entropy as S = A/4. Restoring SI units and substituting κ = c⁴/(4GM) gives T_H = ħc³/(8πGMk_B); with the horizon area A = 16πG²M²/c⁴ the entropy becomes S = 4πk_B GM²/(ħc), or S/k_B = A/(4ℓ_P²) with ℓ_P² = ħG/c³. The radiated power applies the Stefan–Boltzmann law to the horizon, P = σAT_H⁴ with σ = π²k_B⁴/(60ħ³c²), which collapses to P = ħc⁶/(15360πG²M²). Setting the rate of mass loss to −P/c² and integrating gives the lifetime t = 5120πG²M³/(ħc⁴). A quick consistency check: P × t = Mc²/3 exactly. The constants ħ, k_B and c are exact under CODATA 2022; G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻² carries a relative uncertainty of 2.2 × 10⁻⁵, the only real uncertainty in the temperature. Lifetimes are given in Julian years of 31,557,600 s. A mass or temperature of zero is rejected, since the temperature diverges as the mass goes to zero and a zero temperature would need infinite mass.

A worked example.

Example

The example starts from the microwave background's temperature, 2.72548 K, and asks what mass a black hole needs to match it. Inverting the formula, M = ħc³ ÷ (8πGk_BT). The numerator is 1.0546 × 10⁻³⁴ × 2.6944 × 10²⁵ = 2.8414 × 10⁻⁹; the denominator is 8π × 6.6743 × 10⁻¹¹ × 1.3806 × 10⁻²³ × 2.72548 = 6.3121 × 10⁻³². Dividing gives 4.5016 × 10²² kg, or 2.2639 × 10⁻⁸ solar masses, about 61% of the Moon's mass. Its horizon radius is 2GM/c² = 6.6859 × 10⁻⁵ m, so the whole hole is about 0.134 mm across. The entropy is S/k_B = 5.3759 × 10⁶¹ (7.4222 × 10³⁸ J/K), the idealised power is 1.7576 × 10⁻¹³ W and the idealised lifetime 2.4315 × 10⁴⁴ years. T_H ÷ T_CMB is exactly 1: this hole emits as much as it absorbs. Anything heavier is colder than the sky and growing; anything lighter is warmer and shrinking.

hawking Temperature K2.725
mass Kg1,988,410,000,000,000,000,000,000,000,000
solve ForfromTemperature

Frequently asked questions.

How cold are the black holes astronomers have actually found?
Unimaginably cold. A one-solar-mass hole sits at 6.1703 × 10⁻⁸ K and a ten-solar-mass hole at a tenth of that. Sagittarius A*, at about 4.3 million solar masses, comes out at 1.4359 × 10⁻¹⁴ K, and M87*, at about 6.5 billion, at 9.4925 × 10⁻¹⁸ K. A solar-mass hole is about 44 million times colder than the microwave background.
Is any black hole evaporating right now?
None that has been observed. The balance point with the microwave background is 4.5016 × 10²² kg, and the lightest known black holes weigh a few solar masses, around a hundred million times more. Those holes absorb more than they radiate and are slowly growing. They could only start losing mass once the expanding universe has cooled its background radiation below their Hawking temperature.
Why does a black hole get hotter as it loses mass?
Because temperature scales as 1/M, a radiating hole has negative heat capacity: losing energy makes it hotter, which makes it radiate faster. Power scales as 1/M², so the final stage is a runaway burst rather than a gentle fade. The semiclassical theory stops being trustworthy as the mass approaches the Planck mass, about 2.18 × 10⁻⁸ kg, where quantum gravity takes over.
What mass of black hole would just be finishing its evaporation today?
On the idealised photon-only formula, a hole of about 1.73 × 10¹¹ kg has a lifetime equal to the 13.787-billion-year age of the universe; it would sit at 7.0944 × 10¹¹ K. Page's fuller 1976 calculation put the threshold for primordial black holes at (5 ± 1) × 10¹¹ kg, roughly three times heavier, because the extra emission channels shorten every lifetime.
What does the entropy figure actually measure?
Jacob Bekenstein argued in 1973 that a black hole must carry entropy proportional to its horizon area or the second law of thermodynamics would fail; Hawking's temperature fixed the constant at one quarter of the area in Planck units. A solar-mass hole has S/k_B = 1.0489 × 10⁷⁷. Because the area grows as M², doubling the mass quadruples the entropy. Which microscopic states this entropy counts is still an open question in quantum gravity.
Does a spinning black hole have a different temperature?
Yes, it is colder. Everything here assumes the Schwarzschild solution, with no spin and no charge. A rotating hole of the same mass has a lower surface gravity, and its temperature falls towards zero as the spin approaches the maximum allowed. Real astrophysical black holes do spin, so for them the figure shown is an upper limit on the temperature.
Has Hawking radiation ever been detected?
Not from a real black hole, and the numbers show why: a solar-mass hole radiates about 9.0082 × 10⁻²⁹ W while absorbing far more from the microwave background. Laboratory analogues, such as sound waves in Bose–Einstein condensates arranged to form a horizon, have reported thermal emission matching the analogous prediction, but they test the mathematics rather than gravity itself.

References& sources.

  1. [1]Hawking, S. W. (1975), "Particle Creation by Black Holes", Communications in Mathematical Physics 43, 199–220. The original derivation of the thermal spectrum at temperature ħκ/(2πk_B). Peer-reviewed; the journal copy is paywalled at Project Euclid/Springer — cited here as the origin of the result, with the formulas themselves verified against the open-access Wald review below. Retrieved (metadata) 2026-07-29.
  2. [2]Wald, R. M. (2001), "The Thermodynamics of Black Holes", Living Reviews in Relativity 4, 6; open-access full text retrieved 2026-07-29 via PubMed Central. States, in units with G = c = ħ = k = 1, that a black hole "radiates to infinity all species of particles with a perfect black body spectrum, at temperature" T = κ/(2π), and that the entropy is S_bh = A/4 in Planck units. This is the second, independent authority: restoring SI units from those two statements alone reproduces this calculator's T_H and S exactly.
  3. [3]Page, D. N. (1976), "Particle emission rates from a black hole: Massless particles from an uncharged, nonrotating hole", Physical Review D 13, 198. Finds that a hole of mass M ≫ 10¹⁷ g emits a total power of 2 × 10⁻⁴ ħc⁶G⁻²M⁻², of which 81% is neutrinos, 17% photons and 2% gravitons, and that primordial black holes would have decayed within the present age of the universe if and only if their initial mass was below (5 ± 1) × 10¹⁴ g. This is the source of the ~9.7× correction stated beside the power and lifetime outputs. Peer-reviewed; paywalled at APS, abstract freely visible.
  4. [4]Bekenstein, J. D. (1973), "Black Holes and Entropy", Physical Review D 7, 2333. The argument that a black hole must carry entropy proportional to horizon area. Peer-reviewed; paywalled at APS. Bibliographic reference — no number is taken from it that is not independently in Wald.
  5. [5]Fixsen, D. J. (2009), "The Temperature of the Cosmic Microwave Background", Astrophysical Journal 707, 916; arXiv:0911.1955, retrieved 2026-07-29. Gives T_CMB = 2.72548 ± 0.00057 K, the value this page compares every Hawking temperature against. The Particle Data Group independently prints 2.7255(6) K. Peer-reviewed; open-access preprint.
  6. [6]NIST/CODATA 2022 fundamental constants, retrieved 2026-07-29: reduced Planck constant ħ = 1.054 571 817 × 10⁻³⁴ J s (exact), Boltzmann constant k_B = 1.380 649 × 10⁻²³ J K⁻¹ (exact), speed of light c = 299 792 458 m s⁻¹ (exact), Newtonian constant of gravitation G = 6.674 30(15) × 10⁻¹¹ m³ kg⁻¹ s⁻² (u_r = 2.2 × 10⁻⁵). Standards body; open access.
  7. [7]Particle Data Group (Workman et al.), Review of Particle Physics, "Astrophysical Constants and Parameters", Table 2.1, revised August 2023; 2024 PDF retrieved 2026-07-29. Source of M☉ = 1.98841(4) × 10³⁰ kg and of the Planck length ℓ_P = 1.616255(18) × 10⁻³⁵ m, which this calculator reproduces from ħ, G and c. Also prints T_CMB = 2.7255(6) K, agreeing with Fixsen. Open access.
  8. [8]Planck Collaboration (2020), "Planck 2018 results. VI. Cosmological parameters", Astronomy & Astrophysics 641, A6. Source of the 13.787 ± 0.020 Gyr age of the universe used only to decide when to flag a hole as one that would already have evaporated. Peer-reviewed; open access.

How this page was produced

Published by
Quanta Calculator
Primary sources
8 cited below
Method
T_H = ħc³ ⁄ (8πGMk_B) · S = 4πk_B GM² ⁄ (ħc) = k_B A c³ ⁄ (4ħG) · P = ħc⁶ ⁄ (15360πG²M²) · t = 5120πG²M³ ⁄ (ħc⁴)
Published
Last verified

Built with AI assistance and verified by automated tests against the cited sources — every worked example on this page is computed by the same code that runs the calculator. How we build and check calculators.

In this category

Embed

Quanta Pro

Paid features are coming later.

  • All 1560 calculators remain free
  • No billing is enabled
Coming soon