formula.expert

Bomb calorimeter: glucose pellet to calorific value

SCH4U Grade 12 Chemistry · Thermochemistry

A 1.200 g pellet of glucose (C₆H₁₂O₆, M = 180.16 g/mol) is pressed, wired for ignition, and sealed into a bomb calorimeter's steel vessel under oxygen. The bomb sits in a stirred pail of water; a calibration burn of benzoic acid established that the bomb, pail and its 2.000 L of water together absorb heat like 2.48 kg of water — the calorimeter constant, quoted as a water equivalent. After firing, the thermometer climbs from 24.10 °C to 25.90 °C. Water's specific heat is 4186 J/(kg·K).

Given
  • m_p = 1.2 g — Mass of the glucose pellet
  • m_eq = 2.48 kg — Calorimeter water equivalent (calibrated)
  • T₁ = 24.1 °C — Temperature before firing
  • T₂ = 25.9 °C — Temperature after firing
  • M = 180.16 g/mol — Molar mass of glucose
  • c_w = 4186 J/(kg·K) — Specific heat of water
Determine
  1. (a)the heat the burn released into the calorimeter
  2. (b)the calorific value of glucose per unit mass
  3. (c)the amount of glucose burned, in moles
  4. (d)the molar heat of combustion of glucose
Step 1 of 4(a) · solve for Heat energy

Everything the burn released, the calorimeter caught: Q = m_eq·c·ΔT with the CALIBRATED 2.48 kg, not the 2.000 kg of pail water alone. Forgetting the bomb and pail's share — the calorimeter constant — is the classic error, and it reads every fuel about 19% too lean.

Rearranged for Q
Q=mcΔTQ = m c \Delta T
Your values, in your units
Q=(2.48 kg) (4,186 J/(kg⋅K)) (1.8 C∘)Q = \left( 2.48\ \text{kg} \right) \, \left( 4{,}186\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 1.8\ \text{C}^{\circ} \right)
Answer
Q=18.686 kJQ = 18.686\ \text{kJ}

Carried onward at full precision, not this rounded figure.

Open the Sensible Heat (Q = mcΔT) solver →

Step 2 of 4(b) · solve for Specific latent heat

A calorific value is an energy per unit mass — the same Q/m shape as a latent heat, so the same solver does the division. The 18.7 kJ spread over just 1.200 g of pellet is what makes the number large: about 15.6 kJ from every gram, the figure printed per 100 g on food labels after the kilo-conversion.

Rearranged for L
L=QmL = \frac{Q}{m}
18.686 kJcarried from step 1
Your values, in your units
L=(18,686.3 J)(1.2 g)L = \frac{\left( 18{,}686.3\ \text{J} \right)}{\left( 1.2\ \text{g} \right)}
Converted to base units
L=(18,686.3 J)(0.0012 kg)L = \frac{\left( 18{,}686.3\ \text{J} \right)}{\left( 0.0012\ \text{kg} \right)}
Answer
L=15.572 MJ/kgL = 15.572\ \text{MJ/kg}

Carried onward at full precision, not this rounded figure.

Open the Latent Heat solver →

Step 3 of 4(c) · solve for Amount of substance

Chemistry counts in moles, so the pellet's 1.200 g becomes n = m/M — about 6.66 mmol. Keep this number at full precision: it is about to sit under a division, where a rounded 0.0067 would shift the final answer by half a percent all by itself.

Rearranged for n
n=mMn = \frac{m}{M}
Your values, in your units
n=(1.2 g)(180.16 g/mol)n = \frac{\left( 1.2\ \text{g} \right)}{\left( 180.16\ \text{g/mol} \right)}
Converted to base units
n=(0.0012 kg)(180.16 g/mol)n = \frac{\left( 0.0012\ \text{kg} \right)}{\left( 180.16\ \text{g/mol} \right)}
Answer
n=6.6607 mmoln = 6.6607\ \text{mmol}

Carried onward at full precision, not this rounded figure.

Open the Moles from Mass (n = m/M) solver →

Step 4 of 4(d) · solve for Molar enthalpy change

ΔH = q/n converts the burn to the chemist's currency: energy per mole of reaction as written, C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. Strictly the tabulated ΔH_c carries a negative sign — combustion is exothermic — and the calorimeter measures its magnitude; dropping that convention in a Hess's-law problem later flips answers wholesale.

Rearranged for ΔH
ΔH=qn\Delta H = \frac{q}{n}
18.686 kJcarried from step 1
6.6607 mmolcarried from step 3
Your values, in your units
ΔH=(18,686.3 J)(0.00666075 mol)\Delta H = \frac{\left( 18{,}686.3\ \text{J} \right)}{\left( 0.00666075\ \text{mol} \right)}
Answer
ΔH=2.8054 MJ/mol\Delta H = 2.8054\ \text{MJ/mol}

Carried onward at full precision, not this rounded figure.

Open the Heat of Reaction solver →

Therefore the pellet released 18.69 kJ into the calorimeter, glucose delivers about 15.6 kJ per gram, and the 6.661 mmol burned put its molar heat of combustion at 2805 kJ/mol — right beside the handbook's 2803 kJ/mol for ΔH_c of glucose.

Why this order

A bomb calorimeter reduces all of thermochemistry to one thermometer: the order of the chain is the order in which that single ΔT is converted into ever more portable currencies. First joules (the calorimeter's own units, via the calibrated water equivalent), then joules per gram (the engineer's and the nutritionist's number), then moles, then joules per mole (the chemist's, ready for Hess's law). The step that separates a right answer from a plausible one is (a): the calorimeter constant. The steel bomb, the pail and the stirrer all warm by the same 1.80 K as the water, and calibration with a standard — benzoic acid, whose heat of combustion is certified — is how their share gets folded into one effective 2.48 kg. Run the arithmetic with the 2.000 kg of visible water and every answer downstream lands 19% low, consistently and convincingly wrong.

The cross-check built into the chain is that parts (b) and (d) must agree through the molar mass: 15.57 kJ/g × 180.16 g/mol = 2805 kJ/mol, the same number part (d) reaches by its own route — one burn, two currencies, one exchange rate. And the magnitudes deserve their moment: 15.6 kJ/g is the 3.7 kcal/g behind the "4 calories per gram of carbohydrate" on every nutrition panel (bomb values run a shade high because a body does not fully oxidize what a bomb does), and 2805 kJ/mol is precisely the energy budget respiration spends 38 ATP at a time. The derived numbers here are this page's own, from its own readings — close to the handbook because the scenario was built honestly, not copied from it.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.