Experiment 14 Heat Effects And Calorimetry Advance Study Assignment Answers

Experiment 14 Heat Effects and Calorimetry Advance Study Assignment Answers

Understanding how heat is exchanged during chemical reactions is a cornerstone of thermochemistry. Experiment 14 in most general‑chemistry laboratory manuals guides students through a coffee‑cup calorimetry investigation that measures the enthalpy change (ΔH) of a neutralization reaction or the specific heat of a metal. The advance‑study assignment prepares learners to interpret raw temperature data, calculate heat flows, and evaluate sources of error. Below is a detailed walk‑through of the typical advance‑study questions, the reasoning behind each answer, and the calculations you will need to perform in the lab.

1. Introduction to the Experiment

The purpose of Experiment 14 is to determine the heat released or absorbed when a known quantity of reactants mixes in an insulated calorimeter. By measuring the temperature change (ΔT) of the solution and knowing the masses and specific heats of the components, you can calculate the heat (q) exchanged using the fundamental calorimetry equation:

[ q = m , c , \Delta T ]

where m is mass, c is specific heat capacity, and ΔT is the temperature change. The experiment also requires you to find the calorimeter constant (C_cal), which accounts for the heat absorbed by the calorimeter itself (styrofoam cup, stirrer, thermometer). Once C_cal is known, the true heat of reaction (q_rxn) is obtained by adding the calorimeter’s contribution:

[ q_{\text{rxn}} = -(q_{\text{solution}} + q_{\text{calorimeter}}) ]

The negative sign indicates that heat released by the reaction raises the temperature of the surroundings.

2. Theoretical Background

2.1 Heat Capacity vs. Specific Heat

• Specific heat (c) is an intensive property (J g⁻¹ °C⁻¹) that tells how much energy is needed to raise 1 g of a substance by 1 °C.
• Heat capacity (C) is an extensive property (J °C⁻¹) that depends on the amount of material: (C = m \times c).

In calorimetry we often need the heat capacity of the calorimeter (C_cal) because the styrofoam cup, stir bar, and thermometer absorb a non‑negligible amount of heat.

2.2 Enthalpy Change (ΔH)

For reactions carried out at constant pressure (the usual coffee‑cup setup), the heat measured equals the change in enthalpy:

[ \Delta H = \frac{q_{\text{rxn}}}{n} ]

where n is the number of moles of the limiting reactant. ΔH is expressed in kJ mol⁻¹; a negative value denotes an exothermic process, while a positive value denotes an endothermic one.

2.3 Assumptions

1. No heat loss to the environment (perfect insulation). 2. The specific heat of the aqueous solution approximates that of pure water (4.18 J g⁻¹ °C⁻¹) unless otherwise stated.
2. The calorimeter does not react chemically with the solution.
3. Temperature readings are taken promptly after mixing to minimize heat exchange with the surroundings.

3. Advance Study Assignment – Typical Questions & Model Answers

Below are the most common advance‑study prompts found in laboratory manuals for Experiment 14, accompanied by step‑by‑step solutions. Replace the placeholder masses, volumes, and temperature changes with the values given in your specific assignment.

Question 1: Calculate the heat absorbed by the solution (q_solution)

Given:

• Volume of HCl solution = 50.0 mL (density ≈ 1.00 g mL⁻¹ → mass = 50.0 g)
• Volume of NaOH solution = 50.0 mL (mass = 50.0 g)
• Initial temperature of both solutions = 22.0 °C
• Final temperature after mixing = 28.5 °C
• Specific heat of the mixture = 4.18 J g⁻¹ °C⁻¹

Solution:

1. Total mass of solution = 50.0 g + 50.0 g = 100.0 g.
2. ΔT = T_final – T_initial = 28.5 °C – 22.0 °C = 6.5 °C.
3. Apply (q = m c \Delta T):

[ q_{\text{solution}} = (100.0\ \text{g})(4.18\ \text{J g}^{-1}\text{°C}^{-1})(6.5\ \text{°C}) = 2717\ \text{J} ]

Rounded to appropriate significant figures: 2.72 kJ (absorbed by the solution).

Question 2: Determine the calorimeter constant (C_cal) using a known reaction

Given:

• A separate trial mixes 5.00 g of solid NaOH with 50.0 mL of water (no acid).
• The dissolution of NaOH is known to release –44.5 kJ mol⁻¹ (exothermic).
• Molar mass of NaOH = 40.00 g mol⁻¹ → moles NaOH = 5.00 g / 40.00 g mol⁻¹ = 0.125 mol.
• Expected heat released = 0.125 mol × (–44.5 kJ mol⁻¹) = –5.56 kJ (the system loses this amount).
• Measured temperature rise for this trial = 4.2 °C.
• Mass of water = 50.0 g (density 1.00 g mL⁻¹). - Specific heat of water = 4.18 J g⁻¹ °C⁻¹.

Solution:

1. Heat absorbed by the water (q_water):

[ q_{\text{water}} = (50.0\ \text{g})(4.18\ \text{J g}^{-1}\text{°C}^{-1})(4.2\ \text{°C}) = 877.8\ \text{J} = 0.878\ \