Preview

Calorimetry

Q: 50 g of water at 80°C is mixed with 50 g at 20°C. What is the final temperature?

T_f = (50×80 + 50×20)/(50+50) = 50°C (equal masses, simple average).

Q: 100 g of solution rises by 5.2°C in a neutralization. What is q? (c = 4.184 J/(g·°C)).

Q = mcΔT = 100 × 4.184 × 5.2 = 2175.7 J ≈ 2.18 kJ.

Q: If the neutralization used 0.050 mol HCl, what is ΔH per mole?

ΔH = -q/n = -2176/0.050 = -43,520 J/mol ≈ -43.5 kJ/mol.

Heat transfer, specific heat, and enthalpy changes

A thermometer can't count joules. It only reports the average jitter of the molecules it touches — and how MUCH that jitter rises when heat arrives depends on how many molecules share the windfall. The same neutralization, the same heat released: in a small cup the temperature leaps, in a big beaker it barely stirs. Heat is the windfall; temperature change is just how loudly a particular crowd cheers. Today you learn to convert between the two — the accountant's skill called calorimetry.

What you'll be able to do

Calculate heat transfer with q = mcΔT and connect measured ΔT to the heat released or absorbed
Explain why the same heat produces a smaller ΔT in a larger mass (ΔT = q/mc), and convert q to molar ΔH
Connect the sign of a solution's temperature change to the exo/endothermic character of a process

Formulas

q = mc\Delta T

Heat transfer: q = heat (J), m = mass (g), c = specific heat capacity (J/(g·°C)), ΔT = temperature change (°C)

q_{\text{lost}} + q_{\text{gained}} = 0

Conservation of energy in a calorimeter: heat lost by hot substance equals heat gained by cold substance

\Delta H_{\text{rxn}} = -\frac{q_{\text{solution}}}{n}

Enthalpy of reaction per mole: negative of heat absorbed by solution divided by moles of limiting reagent

Make a prediction

0.05 mol of NaOH neutralizes HCl in 100 g of water: the temperature climbs 6.7 °C. Now run the same reaction in 200 g of water. What happens to the temperature rise?

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Your prediction

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Answer: About 3.3 °C — same heat, twice the mass to warm

q = mcΔT, rearranged: ΔT = q/(mc). The reaction releases the same q either way (same moles neutralized) — but the second batch spreads that heat over twice the mass, so each degree of jitter is shared by twice as many molecules: ΔT halves to ≈ 3.35 °C. Temperature change is never the heat itself; it's the heat divided by the crowd. That's why calorimetry needs BOTH a thermometer and a balance.

Quiz (0/3)

50 g of water at 80°C is mixed with 50 g at 20°C. What is the final temperature?

100 g of solution rises by 5.2°C in a neutralization. What is q? (c = 4.184 J/(g·°C))

If the neutralization used 0.050 mol HCl, what is ΔH per mole?

You can now

Calculate heat transfer with q = mcΔT and connect measured ΔT to the heat released or absorbed
Explain why the same heat produces a smaller ΔT in a larger mass (ΔT = q/mc), and convert q to molar ΔH
Connect the sign of a solution's temperature change to the exo/endothermic character of a process