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 40.0 gram block of an unknown metal is cooled to -20.0 °C in a laboratory freezer and then immediately dropped into 140.0 grams of 24.0 °C water in an insulated container. If the final temperature of both the block and water is 22.7 °C, what is the specific heat capacity of the metal?

0.385 J/g°C

0.215 J/g°C

0.900 J/g°C

0.450 J/g°C

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Verified step by step guidance

Start by understanding the concept of heat transfer. When two substances at different temperatures are mixed, heat will flow from the hotter substance to the cooler one until thermal equilibrium is reached.

Use the formula for heat transfer: \( q = m \cdot c \cdot \Delta T \), where \( q \) is the heat absorbed or released, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

Calculate the heat lost by the water using its mass, specific heat capacity (4.18 J/g°C for water), and the change in temperature from 24.0 °C to 22.7 °C.

Calculate the heat gained by the metal block using its mass, the unknown specific heat capacity, and the change in temperature from -20.0 °C to 22.7 °C. Set this equal to the heat lost by the water, as the heat gained by the metal is equal to the heat lost by the water.

Solve for the specific heat capacity of the metal by rearranging the equation: \( c_{metal} = \frac{q_{water}}{m_{metal} \cdot \Delta T_{metal}} \). Substitute the known values to find the specific heat capacity.