AP Chemistry Calorimetry and Specific Heat

Using q = mcΔT for heating, mixing, and coffee-cup reactions, with four worked examples.

Almost every calorimetry question is the same equation wearing a different story: something gets hotter, something gets cooler, and energy is conserved between them. The Unit 6 review has one short example; here the four situations you are most likely to meet are worked out in full, from a plain heating calculation to explaining why an experiment's result disagrees with the accepted value.

The q = mcΔT Method for Every Calorimetry Problem

The core equation is q = mcΔT: heat in joules equals mass in grams times specific heat capacity in J/(g·°C) times the temperature change, where ΔT = T_final − T_initial. Keep the sign: positive q means heat absorbed, negative q means heat released. Because a one-degree change in Celsius equals a one-kelvin change, ΔT is the same in either unit.

Decide what gains heat and what loses it. Heat flows from the hotter object to the cooler one, or from the reaction into the surrounding solution.
Calculate q for the object you have complete data for, with the sign included.
Apply conservation of energy. The heat lost by one object equals the heat gained by the other: q_lost = −q_gained.
Convert to what is asked. Solve for c or the final temperature, or divide the heat of the reaction by the moles that reacted to get ΔH in kJ/mol.

Water's specific heat, 4.18 J/(g·°C), is the one constant worth knowing; for other substances the problem supplies c. These equations apply on the sloped parts of a heating curve, where temperature is changing. On a plateau, where a substance melts or boils, q = nΔH takes over instead.

Worked Example 1: How Much Heat Warms a Sample of Water

How much heat is needed to warm 250.0 g of liquid water from 20.0°C to 65.0°C?

ΔT = 65.0 − 20.0 = 45.0°C.
q = mcΔT = (250.0 g)(4.18 J/(g·°C))(45.0°C) = 47,025 J.
Three significant figures are allowed (the 4.18 and 45.0 limit it), so q = 4.70 × 10⁴ J, or 47.0 kJ. The sign is positive because the water absorbed heat. The significant figures rules decide the rounding.

Worked Example 2: Finding the Final Temperature When Hot Metal Meets Cool Water

A 50.0 g piece of aluminum (c = 0.897 J/(g·°C)) at 100.0°C is dropped into 100.0 g of water at 20.0°C in an insulated cup. Find the final temperature of both.

The unknown is the final temperature, T_f, which is the same for both objects at equilibrium.
q_Al = −q_water
(50.0)(0.897)(T_f − 100.0) = −(100.0)(4.18)(T_f − 20.0)
44.85 T_f − 4485 = −418 T_f + 8360
462.85 T_f = 12,845, so T_f = 27.8°C.

Check that the answer is reasonable: it lies between 20.0°C and 100.0°C, and it is much closer to the water's starting temperature because the water's heat capacity (418 J/°C) is about nine times the metal's (44.85 J/°C). Seeing that before you calculate is a quick way to catch a sign error.

Worked Example 3: Finding ΔH of Neutralization in a Coffee-Cup Calorimeter

50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH, both at 21.0°C, are mixed in a coffee-cup calorimeter. The highest temperature reached is 27.8°C. Assume the solution has the density (1.00 g/mL) and specific heat of water and that the cup absorbs no heat. Find ΔH for the reaction in kJ/mol.

Heat gained by the solution: the mass is 100.0 mL × 1.00 g/mL = 100.0 g and ΔT = 27.8 − 21.0 = 6.8°C, so q_solution = (100.0)(4.18)(6.8) = 2842 J, about +2.8 kJ.
Heat of the reaction: q_rxn = −q_solution = −2.8 kJ.
Moles reacted: (0.0500 L)(1.00 mol/L) = 0.0500 mol of each. They are in exact 1:1 ratio, so neither is limiting and 0.0500 mol of water forms.
ΔH: −2842 J / 0.0500 mol = −56,840 J/mol, which rounds to −57 kJ/mol.

The reaction is exothermic, which agrees with the solution warming up. Getting the moles from molarity and volume is the same skill as in the mole conversions guide.

Worked Example 4: Explaining Why an Experimental ΔH Differs From the Accepted Value

A student repeating Example 3 calculates ΔH = −54 kJ/mol, while the accepted value is closer to −57 kJ/mol. Identify a source of error that would produce this result, and explain it.

Some heat escaped from the cup to the room, or warmed the cup itself, instead of warming the solution. Then the measured ΔT is smaller than it should be, which makes q_solution and the calculated heat of the reaction too small in magnitude, so ΔH comes out less negative than the true value. The direction of the error is what a free-response question checks, so state both what was lost and how it changed ΔT.

If the problem gives a heat capacity for the calorimeter itself, include it: the calorimeter's heat is C_cal × ΔT, added to the solution's before changing the sign. Other plausible causes in a question like this include assuming the solution's specific heat or density equals water's when the solution is more concentrated, reading the temperature before the mixture has reached its maximum, and imprecise volume or concentration measurements.

Common Calorimetry Mistakes

Dropping the sign. Calculate ΔT as final minus initial and keep it; a negative q for the metal is what makes the energy balance work.
Forgetting to flip the sign for the reaction. q_rxn = −q_solution. A warmer solution means a negative ΔH.
Using the mass of the water only. In a mixing problem, each object gets its own m, c, and ΔT.
Reporting J/mol where kJ/mol is expected. Divide by 1000 at the end, and use moles of the reaction as written, not grams.
Using q = mcΔT on a plateau. During a phase change ΔT is zero; use q = nΔH_fus or nΔH_vap.

The heat values from calorimetry feed straight into larger thermochemistry problems such as Hess's law calculations. You can find every other free guide and scoring tool starting from the AP Chem Score Calculator, which is also where to estimate how your practice results would translate to an exam score.

Frequently Asked Questions

What is specific heat capacity and what are its units?

Specific heat capacity (c) is the heat needed to raise the temperature of one gram of a substance by one degree Celsius (or one kelvin). Its units are J/(g·°C), equivalently J/(g·K). For liquid water it is 4.18 J/(g·°C).

What is the formula for heat in a calorimetry problem?

q = mcΔT, where q is the heat absorbed or released in joules, m is the mass in grams, c is the specific heat capacity, and ΔT is the final temperature minus the initial temperature. A positive q means the substance absorbed heat; a negative q means it released heat.

Why is the heat of the reaction the negative of the heat of the solution?

Energy is conserved. In a coffee-cup calorimeter the solution gains exactly the heat the reaction releases, so q(reaction) = −q(solution). A reaction that warms the solution is exothermic, so its q and ΔH are negative.

Does it matter if I use Celsius or kelvins for ΔT?

No. A change of one degree Celsius equals a change of one kelvin, so ΔT has the same value in either unit. You only need kelvins when a formula uses the temperature itself, not a difference.

What if the problem gives the heat capacity of the calorimeter?

Then the calorimeter absorbs some of the heat as well. Calculate its heat as C(cal) × ΔT and add it to the heat gained by the solution before taking the negative to find the heat of the reaction. If the problem does not mention it, the usual assumption is that the cup absorbs none.

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