This walks through the calculator’s built-in default conditions step by step, showing where the headline value of Q = 62.37 kJ comes from and confirming it agrees with the literature for industrial anaerobic ethanol fermentation. To reproduce, click Reset to defaults on the Calculator tab.
Step 1 — Inputs
Glucose mass = 100 g
Mass yields (% w/w of glucose):
Ethanol = 46 % → Yeth = 0.46
Glycerol = 4 % → Ygly = 0.04
Acetic acid = 1 % → Yace = 0.01
Yeast biomass = 2 % → Ybio = 0.02
Σ = 53 % (47 % to CO₂ + H₂O by mass)
Heats of combustion (HHV at 25 °C, kJ/g):
Glucose = 15.56
Ethanol = 29.67
Glycerol = 17.96
Acetic acid = 14.57
Yeast biomass = 21.20
Step 2 — Energy entering as glucose
Treat the glucose feed as a chemical fuel and compute its full combustion potential:
Eglucose = mglucose × ΔH°c,glucose
= 100 g × 15.56 kJ/g
= 1556.00 kJ
Step 3 — Energy retained in each product
Each named product carries its own combustion potential out of the system. Compute one row per product:
Eethanol = Yeth × m × ΔH°c,eth
= 0.46 × 100 × 29.67 = 1364.82 kJ
Eglycerol = Ygly × m × ΔH°c,gly
= 0.04 × 100 × 17.96 = 71.84 kJ
Eacetic = Yace × m × ΔH°c,ace
= 0.01 × 100 × 14.57 = 14.57 kJ
Ebiomass = Ybio × m × ΔH°c,bio
= 0.02 × 100 × 21.20 = 42.40 kJ
Σ retained = 1493.63 kJ
CO2 and H2O are already at the fully-oxidised reference state of the HHV scale, so their combustion contribution is zero and they drop out of the sum.
Step 4 — Heat released by difference
Q = Eglucose − Σ retained
= 1556.00 − 1493.63
= 62.37 kJ (heat released to the broth)
ΔHrxn = −Q
= −112.4 kJ / mol glucose (chemistry sign convention, exothermic)
Step 5 — Carbon balance closure
An independent check: trace carbon atoms instead of energy. They must conserve.
nC(glucose) = 100 × 6 / 180.16 = 3.330 mol C
nC(ethanol) = 0.46 × 100 × 2 / 46.07 = 1.997 mol C
nC(glycerol) = 0.04 × 100 × 3 / 92.09 = 0.130 mol C
nC(acetic) = 0.01 × 100 × 2 / 60.05 = 0.033 mol C
nC(biomass) = 0.02 × 100 / 24.626 = 0.081 mol C
Σ in products = 2.242 mol C (67.3 %)
nC(CO₂) = nC(glucose) − Σ products = 1.089 mol C
mass CO₂ = 1.089 × 44.01 = 47.91 g CO₂
Step 6 — Cross-check against the literature
Industrial anaerobic ethanol fermentation with normal side-product yields is reported in several independent sources:
- Heijnen & van Dijken (1992), thermodynamic ethanol-fermentation calculations: ~96 kJ/mol glucose with biomass formation.
- Roels (1983), Energetics & Kinetics in Biotechnology: ~84 kJ/mol glucose for glucose → ethanol + biomass.
- Industrial brewing & ethanol-distillery handbooks: 130–150 kcal/kg sugar ≈ 54–63 kJ per 100 g glucose.
- This calculator at defaults: 112 kJ/mol glucose ≡ 62.4 kJ per 100 g.
The result sits at the upper end of the cited range — consistent with the moderate biomass-yield assumption. It is not the cooling load of an industrial fermenter; that quantity is typically 2–3× higher because it includes agitator dissipation, sparge-gas heating, microbial-maintenance work, and thermal infiltration, none of which are part of the Hess balance.
Step 7 — Why so much glucose energy stays out of Q
Of the 1556 kJ of combustion potential entering as glucose, only 62 kJ (4 %) is released as heat. The remaining 96 % is locked into the chemical bonds of ethanol, glycerol, acetate, and biomass — products that are themselves combustible fuels. To liberate that energy you would have to oxidise the products in a downstream step (combustion of distillate, biomass digestion, etc.); it is not available to warm the broth.
As a contrast, fully-aerobic propagation with all carbon committed to biomass (Ybio ≈ 50 %, others 0) gives Q ≈ 496 kJ per 100 g glucose — about 8× larger than this anaerobic-ethanol case. Try those numbers in the calculator to see why aerobic propagation tanks need much heavier cooling.