Three kinds of “heat capacity”
- Specific heat capacity (c)比热容 is the heat needed to raise 1 g of a substance by 1 °C, in J/(g·°C). It is a property of the substance; for water it is 4.18 J/(g·°C).
- Heat capacity (C)热容 is the heat needed to raise a whole object by 1 °C: C = m·c, in J/°C. It depends on how big the object is.
- Molar heat capacity is the heat needed to raise 1 mol of a substance by 1 °C, in J/(mol·°C); it equals the specific heat × the molar mass.
All three fit the same relationship: q = m·c·ΔT = C·ΔT. The AP equation sheet gives q = mcΔT but not the specific heat of water, which questions supply.
The coffee-cup calorimeter
For a reaction at constant pressure (an open foam cup), the heat released or absorbed equals ΔH. The foam cup is insulated, so all of the reaction’s heat can be taken as absorbed by, or supplied by, the solution:
Dividing by the moles that react gives ΔH (kJ/mol):
Watch for these points:
- Sign: if the solution warms (ΔT > 0), q(soln) > 0, so the reaction is exothermic and ΔH < 0.
- Use the limiting reactant for n: excess reactant is left over and releases no heat.
- Scale: if both solution volumes double, the amount reacting and the solution’s mass both double, so ΔT stays the same, the heat released doubles, and ΔH (per mole) is unchanged.
What data are needed?
To measure ΔH by calorimetry, you need at least: the mass of the solution (or its volume and density), its specific heat, the initial and highest (or lowest) temperatures, and the moles of the limiting reactant. Data such as atmospheric pressure, boiling point, or heat of fusion aren’t needed.
Error analysis
Error-analysis questions usually ask whether a mistake makes the calculated ΔH too large or too small. First work out how the mistake affects the measured ΔT (or m, or n), then follow it through the equations. The lab below switches to an endothermic dissolving experiment (NH₄NO₃ in water, which gets colder). Choose a mistake to see how the student’s data and calculation change:
8.00 g of solid NH₄NO₃ dissolves in 100.0 g of water in a coffee-cup calorimeter, starting at 22.0 °C; dissolving absorbs heat, so the solution gets colder. The student calculates with a solution mass of 108.0 g, a specific heat of 4.18 J/(g·°C), and n(NH₄NO₃) = 8.00 ÷ 80.05 = 0.0999 mol.
The sizes of the mistakes are schematic: the metal cup lets the solution cool only 85% as much, and the real specific heat is taken as 3.90 J/(g·°C). The enthalpy of solution, +25.69 kJ/mol, is from the CRC Handbook (infinite dilution).
Example 1: ΔH from dissolving data
4.00 g of solid NaOH is dissolved in 96.0 g of water, and the temperature rises from 21.0 °C to 31.6 °C. Taking the solution’s specific heat as 4.18 J/(g·°C), find the enthalpy of solution of NaOH.
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Heat absorbed by the solution. The solution’s mass is NaOH plus water, and the temperature rises by 31.6 − 21.0 = 10.6 °C (heat in J):
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Heat released by dissolving, by conservation of energy. The heat the solution gains comes from the dissolving process, so the signs are opposite (in J, that is, −4.43 kJ):
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Moles of NaOH.
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Enthalpy of solution. ΔH is the heat per mole:
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Check. The negative sign matches “the solution warmed, so heat was released”; the value is close to the CRC Handbook’s −44.51 kJ/mol, with the difference coming from heat loss and from taking the solution’s specific heat as water’s.
Example 2: What happens when the volumes double?
Trial 1: 50.0 mL of 1.0 M HCl is mixed with 50.0 mL of 1.0 M NaOH; the temperature rises by ΔT₁ and heat X is released. Trial 2: both solutions are 100.0 mL, at the same concentrations. Compare the two trials’ ΔT, heat released, and calculated ΔH.
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Compare the amount reacting. Same concentrations, double the volumes: the moles of HCl and NaOH both double, twice as much water forms, and the heat released is 2X.
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Compare the solution mass. Trial 2 has 200.0 mL of solution in all, twice trial 1, so twice the mass.
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Compare ΔT. From q = mcΔT:
Heat and mass both double, so ΔT is unchanged.
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Compare ΔH. ΔH is the heat per mole of reaction:
Both trials give the same ΔH.
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Summary.
- Heat released: doubled (extensive)
- ΔT: unchanged
- ΔH (per mole): unchanged
- The foam cup is replaced by a metal cup that conducts heat away
ΔT is too small, so the calculated heat is too small.
- The thermometer has a zero error, so every reading is 1.2 °C below the true temperature
The two readings are subtracted, so the error cancels.
- The highest temperature is read while it is still rising
ΔT is too small.
- Only the water’s mass is used, without adding the solute’s mass
m is too small, so q is too small.
- Dividing by the moles of the excess reactant (more than the limiting reactant)
n is too large, so the heat per mole is too small.
- The solution’s true specific heat is less than 4.18, but 4.18 is still used
Less heat was actually released than calculated, so the result is too large.
- The heat absorbed by the calorimeter itself is ignored
The solution warms less, so ΔT is too small.
- Some of the solid solute doesn’t dissolve, but the calculation assumes it all did
Less heat is released, yet n counts all of it.
- Both solution volumes are doubled (same concentrations)
ΔT and ΔH (per mole) don’t change.
Common mistakes
Mistake 1: “q is positive, so the reaction is endothermic.” First be clear whose q it is. When the solution warms, q(soln) > 0; that heat came from the reaction, so q(rxn) < 0 and the reaction is exothermic.
Mistake 2: “If both solutions are doubled, the temperature rise doubles.” The heat released doubles, but so does the solution to be heated, so ΔT stays the same.
Mistake 3: “Heat capacity and specific heat are the same thing.” Specific heat is a per-gram property of a substance; heat capacity belongs to a whole object, C = m·c. 100 g of water has 10 times the heat capacity of 10 g, but the same specific heat.
Quick check
25.0 mL of 2.00 M HCl is mixed with 25.0 mL of 2.00 M NaOH, and the temperature rises from 20.0 °C to 33.4 °C. The solution’s mass is 50.0 g and c = 4.18 J/(g·°C). About what is ΔH for the neutralization?
q(soln) = 50.0 × 4.18 × 13.4 = 2.80 × 10³ J; n = 0.0500 mol; ΔH = −2.80 kJ ÷ 0.0500 mol = −56 kJ/mol. A doesn’t divide by n; C has the wrong sign.
Trial 1 uses 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH and releases heat X; trial 2 uses 50 mL of 1.0 M HCl and 100 mL of 1.0 M NaOH and releases heat Y. How are X and Y related?
In trial 2 the NaOH is in excess, so still 0.050 mol of water forms and the same heat is released; there is just more solution, so ΔT is smaller.
In the lab above, 8.00 g of NH₄NO₃ dissolved in 100.0 g of water lowers the temperature by about 5.7 °C. To make a cold pack that lowers the temperature of 100.0 g of water by about 11 °C, roughly how much NH₄NO₃ is needed?
The heat absorbed is proportional to the amount of NH₄NO₃ that dissolves. With the same mass of water, roughly doubling the temperature drop takes roughly twice as much NH₄NO₃. (The solution mass goes from 108 g to 116 g, so the actual drop is about 10.6 °C.)
To determine a reaction’s ΔH by calorimetry, what must be known besides the solution’s mass, the temperature change, and the moles of the limiting reactant?
q = m·c·ΔT needs the specific heat; the other data play no part in the calculation.
A bomb calorimeter (a sealed steel vessel in water) has a total heat capacity of 10.0 kJ/°C. Burning 0.500 g of sucrose in it raises the temperature by 0.825 °C. How much energy is released per gram of sucrose?
The heat capacity is for the whole calorimeter: q = C·ΔT = 10.0 × 0.825 = 8.25 kJ (B is the total for 0.500 g), so per gram 8.25 ÷ 0.500 = 16.5 kJ. C multiplies by the mass; D also multiplies by the specific heat of water, but C already covers the whole calorimeter.
In a neutralization calorimetry experiment, which change makes the temperature rise more without changing the calculated ΔH?
Doubling the concentrations at the same volumes doubles the amount of reaction while the mass of solution to be heated stays about the same, so ΔT roughly doubles; the heat per mole is unchanged, so ΔH is unchanged. A leaves ΔT the same; C and D both make ΔT smaller, and C also makes the calculated ΔH too small.
50.0 g of water warms from 21.6 °C to 28.4 °C, and c = 4.18 J/(g·°C). With the correct significant figures, how much heat did the water absorb?
ΔT = 6.8 °C has only two significant figures, so q = 50.0 × 4.18 × 6.8 = 1.4 × 10³ J.
AP-style free response
A student mixes 50.0 mL of 0.500 M HCl with 75.0 mL of 0.500 M NaOH in a foam cup; both start at 22.0 °C, and the highest temperature is 24.7 °C. Take the solution’s density as 1.00 g/mL and its specific heat as 4.18 J/(g·°C).
(a) Which reactant is limiting? How many moles of water form?
(b) Calculate the heat absorbed by the solution.
(c) Calculate ΔH for the neutralization in kJ/mol, with the correct sign and significant figures.
(d) If 100.0 mL of 0.500 M NaOH were used instead (with the same HCl), predict the temperature rise, and explain.
(e) If the student used only the 50.0 g of the HCl solution as the solution mass, would the magnitude of ΔH calculated in (c) be too large or too small?
Model answer and scoring
(a) n(HCl) = 0.0250 mol and n(NaOH) = 0.0375 mol, so HCl is limiting, and 0.0250 mol of water forms.
(b) m = 125.0 g; q(soln) = 125.0 g × 4.18 J/(g·°C) × 2.7 °C = 1.4 × 10³ J (1411 J).
(c) ΔH = −1.41 kJ ÷ 0.0250 mol = −56 kJ/mol (ΔT has only two significant figures).
(d) Still 0.0250 mol of water forms, so the same heat (1411 J) is released, but the solution is now 150.0 g: ΔT = 1411 J ÷ (150.0 g × 4.18 J/(g·°C)) = 2.25 °C, or about 2.3 °C, smaller than before (or by proportion: 2.7 °C × 125.0 ÷ 150.0).
(e) Too small. All 125.0 g of solution absorbed the heat, but only 50.0 g was used for m, so the calculated heat released is only 2/5 of the actual heat, and the magnitude of ΔH is too small.
Scoring: (a) limiting reactant and n; (b) mass includes both solutions; (c) sign, dividing by n, and significant figures; (d) same heat, larger mass; (e) too small, with a reason (m is too small).
Summary
- Specific heat c (per gram), heat capacity C = m·c (a whole object), and molar heat capacity (per mole); q = m·c·ΔT.
- Coffee-cup calorimeter: q(rxn) = −m(soln)·c·ΔT, and ΔH = q(rxn) ÷ n, with n from the limiting reactant. A warming solution means an exothermic reaction with ΔH < 0.
- If both solution volumes double, the heat doubles while ΔT and ΔH (per mole) stay the same.
- Error analysis: first see how the mistake affects ΔT, m, or n, then decide whether the result is too large or too small; the same offset in both readings cancels out.
AP tip: In a calculation, say whether q refers to the solution or the reaction, and give the final ΔH with its sign and units of kJ/mol. Watch significant figures: the temperature difference often limits the result to two or three.