Energy per Operation

Every irreversible bit erasure has a temperature-dependent minimum energy; real events sit above it.

Worked example — Energy per Operation

Selected scale	one femtojoule
Minimum irreversible bit erasure	2.967 × 10⁻²¹ J/bit
Comparison energy	1.000 × 10⁻¹⁵ J/event
Multiple of Landauer bound	3.37 × 10⁵×
Energy-only throughput ceiling	2.00 × 10¹⁶ events/s

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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The floor belongs to erasure, not every logic step 🖖

Landauer's kT ln 2 cost applies when logically irreversible processing destroys one bit of information. Reversible transformations can in principle avoid that particular minimum, though real machines still dissipate energy.

Cooling lowers the floor and raises the bill 🖖

kT ln 2 is linear in temperature, so going from 310 K to 77 K lowers the floor by 4.03×. But even a perfect refrigerator needs 2.9 J of work to move each joule of waste heat from 77 K into a 300 K room, so the total energy per erasure improves by a factor of 1.03. The floor moves a long way. The bill barely moves.

Erasing a gram of DNA costs seven joules 🖖

A gram of DNA holds about 2.9 × 10²⁰ bytes, which is 2.3 × 10²¹ bits. Erasing every one of them at the 310 K floor costs 7 J — under a tenth of a percent of an AA cell. Landauer’s bound is not what makes computing expensive. The factor of 337,000 that real silicon sits above it is.

Problem solved in full

A phone battery that lasts 10,000 hours, and the same battery lasting 384,500 years 6 steps
Work out the least energy physics allows for erasing one bit at body temperature, compare it with what a silicon logic event actually costs, and then spend a 10 Wh phone battery both ways at a million million operations a second.
1. Landauer's bound says erasing one bit of information costs at least k_BT ln 2, and no rearrangement of the circuit avoids it, because the cost is for destroying the information rather than for moving the charge.
  
  $E_{m i n} = k_{B} T ln 2 = 1.380649 \times 1 0^{- 23} \times 310 \times 0.69315 = 2.967 \times 1 0^{- 21} J$
2. Put body temperature in. 1.380649 × 10⁻²³ × 310 × 0.69315 = 2.967 × 10⁻²¹ J per bit. That is the whole floor, and it is a very small number.
3. A logic event on current silicon costs about 1 × 10⁻¹⁵ J. Divide: 10⁻¹⁵ ÷ 2.967 × 10⁻²¹ = 337,077. Real hardware runs about 337,000 times above the floor.
  
  $\frac{1 0 ^{- 15}}{2.967 \times 1 0 ^{- 21}} = 337, 077$
4. Now spend a battery. 10 Wh is 10 × 3600 = 36,000 J. At the real cost that buys 36,000 ÷ 10⁻¹⁵ = 3.60 × 10¹⁹ operations; at the floor it would buy 36,000 ÷ 2.967 × 10⁻²¹ = 1.21 × 10²⁵.
  
  $N_{real} = \frac{36 , 000}{1 0 ^{- 15}} = 3.60 \times 1 0^{19}, N_{m i n} = \frac{36 , 000}{2.967 \times 1 0 ^{- 21}} = 1.21 \times 1 0^{25}$
5. Run them at 10¹² operations per second. The real battery lasts 3.60 × 10⁷ s, which is 10,000 hours. The floor-limited one lasts 1.21 × 10¹³ s, which is 384,500 years.
  
  $\frac{3.60 \times 1 0 ^{19}}{1 0 ^{12}} = 10, 000 h, \frac{1.21 \times 1 0 ^{25}}{1 0 ^{12}} = 384, 500 yr$
6. Try to close the gap by cooling. Dropping 310 K to 300 K lowers the floor to 2.871 × 10⁻²¹ J, a saving of 3.2%, because the bound is linear in T. The factor of 337,077 is not a temperature effect and no thermostat touches it.
  
  $k_{B} (300) ln 2 = 2.871 \times 1 0^{- 21} J \Rightarrow 3.2% lower$
Answer

The same battery, at the same operation rate, lasts 10,000 hours in silicon and 384,500 years at the thermodynamic limit. Which settles what Landauer's bound is for. It is not the reason computers need power, and it is not a target anyone is approaching: at 337,077 times the floor, essentially all of a chip's energy is going into charging and discharging capacitance, driving wires and leaking, and none of it is the irreducible cost of forgetting. So a page that says "computing has a fundamental energy cost" is true and almost entirely beside the point. The number worth quoting is the ratio, not the bound. And note what the bound does not charge for: a reversible operation erases nothing, so it has no floor at all — which is the whole reason reversible computing is a research field rather than a curiosity.

Learning path

Computing with molecules

References (2)

The original statement of the irreversible-computation energy bound: R. Landauer, “Irreversibility and Heat Generation in the Computing Process.” IBM Journal of Research and Development 5(3), 183–191, 1961.
Direct experimental approach to the kT ln 2 limit in a one-bit memory: A. Bérut et al., “Experimental verification of Landauer’s principle linking information and thermodynamics.” Nature 483, 187–189, 2012.

By the 42Math team · Updated 2026-08-22

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Landauer bound - The floor itself: 2.967 × 10⁻²¹ J per bit erased at body temperature.
ATP-scale: 20 kT - One ATP hydrolysis is 2.89 × 10¹ times the floor — biology computes within thirtyfold of thermodynamics.
one femtojoule - A femtojoule switch sits 3.37 × 10⁵ times above the same floor.
one picojoule - At a picojoule per event, a 20 W budget buys 2.00 × 10¹³ events per second.