AP Chemistry · Topic 6.4
Heat Capacity and Calorimetry Practice
Part of Thermodynamics.(ENE-2.D)
Practice questions
26
Sample questions
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Sample 1difficulty 1/5
To determine the heat of neutralization, a student mixes 50.0 mL of 1.00 M HCl with 50.0 mL of 1.00 M NaOH in a polystyrene calorimeter. Initial temperature of both solutions = 21.5 C. Maximum temperature after mixing = 28.2 C. Assume density 1.00 g/mL and specific heat 4.18 J/(g*C). Heat capacity of the calorimeter is negligible.
What is the molar enthalpy of neutralization (delta H) for HCl + NaOH?
- A
-112 kJ/mol
- Bcheck_circle
-56.0 kJ/mol
- C
-2.80 kJ/mol
- D
+56.0 kJ/mol
Why
q = m<em>c</em>deltaT = 100.0 * 4.18 * 6.7 = 2801 J released. Moles H2O formed = 0.0500 mol. delta H = -2801/0.0500 = -56,020 J/mol = -56.0 kJ/mol.
- A
Sample 2difficulty 2/5
A 1.500 g sample of benzoic acid (deltaU_combustion = -26.43 kJ/g) is burned in a bomb calorimeter, raising the temperature of the bomb + water from 23.55 C to 27.32 C. In a second experiment, a 0.952 g sample of an unknown solid fuel is burned in the same calorimeter, and the temperature rises by 2.41 C.
What is the heat of combustion per gram of the unknown fuel?
- A
-12.5 kJ/g
- Bcheck_circle
-26.6 kJ/g
- C
-2.41 kJ/g
- D
-39.65 kJ/g
Why
q = C_cal * deltaT = 10.51 * 2.41 = 25.33 kJ released. Per gram: 25.33 / 0.952 = 26.6 kJ/g; sign negative for combustion.
- A
Sample 3difficulty 2/5
A student determines the specific heat of an unknown metal. A 50.00 g chunk of metal is heated in a boiling water bath to 99.5 C and quickly transferred into 100.0 g of water at 22.0 C in a coffee-cup calorimeter. The water temperature rises to a maximum of 25.0 C. Specific heat of water = 4.18 J/(g*C). Calorimeter heat capacity is negligible.
What is the specific heat of the metal?
- A
0.067 J/(g*C)
- B
4.18 J/(g*C)
- Ccheck_circle
0.337 J/(g*C)
- D
0.840 J/(g*C)
Why
q_water = 100.0 * 4.18 * 3.0 = 1254 J. q_metal = -1254 J. c_metal = 1254 / (50.00 * (99.5-25.0)) = 1254 / 3725 = 0.337 J/(g*C).
- A
Sample 4difficulty 2/5
To determine the heat of neutralization, a student mixes 50.0 mL of 1.00 M HCl with 50.0 mL of 1.00 M NaOH in a polystyrene calorimeter. Initial temperature of both solutions = 21.5 C. Maximum temperature after mixing = 28.2 C. Assume density 1.00 g/mL and specific heat 4.18 J/(g*C). Heat capacity of the calorimeter is negligible.
If instead 100.0 mL of 1.00 M HCl were mixed with 100.0 mL of 1.00 M NaOH (same initial T), the predicted maximum temperature change is approximately:
- A
halved, about 3.4 C
- B
zero
- C
double, about 13.4 C
- Dcheck_circle
the same 6.7 C
Why
Doubling reactant amounts doubles q released and doubles the mass of solution; q/(m*c) gives the same delta T (intensive temperature change).
- A
Sample 5difficulty 2/5
A coffee-cup calorimeter consists of an insulated polystyrene cup containing water and a thermometer, open to the air.
Which assumption is fundamental to using a coffee-cup calorimeter to measure ΔH?
- A
The water boils during the reaction
- B
No heat is absorbed by the water
- C
The reaction occurs at constant volume
- Dcheck_circle
The reaction occurs at constant atmospheric pressure
Why
A coffee-cup calorimeter is open to the atmosphere, so q_p = ΔH (constant-pressure calorimetry).
- A