Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 36m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

8. Thermochemistry

Heat Capacity

General Chemistry

8. Thermochemistry

Heat Capacity

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Multiple Choice

A volume of 75.0 mL of H2O is initially at room temperature (22.00 °C). A chilled steel rod at 2.00 °C is placed in the water. If the final temperature of the system is 21.10 °C, what is the mass of the steel bar? Use the following specific heat capacities: c_water = 4.18 J/g°C, c_steel = 0.452 J/g°C.

250 g

150 g

300 g

200 g

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Verified step by step guidance

Identify the principle of conservation of energy, which states that the heat lost by the water will be equal to the heat gained by the steel rod. This can be expressed as: \( q_{water} = -q_{steel} \).

Calculate the heat lost by the water using the formula: \( q = m \cdot c \cdot \Delta T \), where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. First, find the mass of the water: \( m_{water} = \text{density of water} \times \text{volume} = 1.00 \text{ g/mL} \times 75.0 \text{ mL} \).

Determine the change in temperature for the water: \( \Delta T_{water} = T_{final} - T_{initial} = 21.10 \text{ °C} - 22.00 \text{ °C} \).

Set up the equation for the heat gained by the steel rod: \( q_{steel} = m_{steel} \cdot c_{steel} \cdot \Delta T_{steel} \), where \( \Delta T_{steel} = T_{final} - T_{initial} = 21.10 \text{ °C} - 2.00 \text{ °C} \).

Solve for the mass of the steel rod \( m_{steel} \) by equating the heat lost by the water to the heat gained by the steel: \( m_{water} \cdot c_{water} \cdot \Delta T_{water} = m_{steel} \cdot c_{steel} \cdot \Delta T_{steel} \). Rearrange the equation to solve for \( m_{steel} \).