Thermochemistry

the idea

Temperature is how hot. Heat is how much energy moved.

A cup of 72 °C milk and a 6,000-gallon silo at 72 °C are the same temperature, but the silo holds vastly more thermal energy. Heat (q) depends on how much stuff (m), how stubborn it is about changing temperature (c, the specific heat), and how far the temperature moved (ΔT).

q = m · c · ΔT

ΔT = T_final − T_initial · c for water = 4.184 J/(g·°C)

Plant version: specific heat is why the CIP water tank takes forever to heat but stainless pipe gets hot to the touch almost immediately. Water's c is about 8× stainless steel's (4.18 vs 0.50 J/g·°C), so every gram of water needs 8× the energy for the same temperature rise.

Exothermic: ΔH < 0The system releases heat to the surroundings, which warm up. Combustion, neutralization, freezing, condensing steam.

Endothermic: ΔH > 0The system absorbs heat from the surroundings, which cool down. Melting ice, evaporation, cold packs.

Sign convention: everything is from the system's point of view, like a bank account. Negative means energy left the system.

calculate

q = mcΔT

Material Mass (g)

Initial temp (°C) Final temp (°C) c (J/g·°C)

calorimetry

Drop something hot into water

Heat lost by the hot object = heat gained by the water. They meet at one final temperature: q_hot + q_cold = 0.

Hot object Its mass (g) Its temp (°C)

Water mass (g) Water temp (°C)

engineering version

The regen section as an energy balance

Hot pasteurized milk heats incoming cold raw milk, and the raw milk cools the pasteurized milk. Whatever energy the regen recovers is energy the heating section doesn't have to supply and the cooling section doesn't have to remove. It's q = mcΔT run per second, which gives you power.

Flow (gal/min) Raw milk in (°C) Pasteurize at (°C)

Regeneration efficiency:

Assumes whole milk at c ≈ 3.93 kJ/(kg·°C) and density 1.03 kg/L, and that the outgoing milk is cooled back to the raw-milk temperature. Real plates lose a little heat to the room. This is the idealized balance.

enthalpy

Reaction heat scales with moles, and Hess's law adds steps

ΔH in a thermochemical equation is per the moles written. CH₄ + 2O₂ → CO₂ + 2H₂O, ΔH = −890 kJ, means 890 kJ released per 1 mol of CH₄ burned. Burn 2 mol and you get 1,780 kJ. So ΔH works as one more conversion factor on the mole map.

Hess's law rules

Energy is a state function: only the start and end matter, not the route.
• Flip a reaction → flip the sign of ΔH
• Multiply a reaction by n → multiply ΔH by n
• Add reactions → add ΔH values

Worked example

C + ½O₂ → CO ΔH = −110.5 kJ
CO + ½O₂ → CO₂ ΔH = −283.0 kJ
Sum: C + O₂ → CO₂ ΔH = −393.5 kJ
CO cancels because it's made in step 1 and used in step 2.

practice

Heat problems

exam traps

Common mistakes

Computing ΔT as initial − final. It's always final − initial, and the sign is the answer to "did it gain or lose heat."
Mixing J and kJ, or g and kg, with a c value in other units. Match c's units.
Forgetting the negative sign in calorimetry: q_metal = −q_water.
Using q = mcΔT across a phase change. Melting and boiling happen at constant T and need ΔH_fus or ΔH_vap, not c.