What is the final temperature of gold. The specific heat capacity is 0.128 j/g. C

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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What is the final temperature of gold. The specific heat capacity is 0.128 j/g. C
This image depicts a multiple-choice question commonly found in educational materials, particularly in subjects like chemistry or physics, where temperature measurements are relevant. Below are four options listed as possible answers:

1. 75.9°C
2. 51.1°C
3. 58.1°C
4. 101°C

Each option is preceded by a radio button, indicating that this format is likely part of a digital quiz or an online test. This type of question format is typically used to assess a student's understanding of a concept or the correct interpretation of data. There are no graphs or diagrams accompanying this question.
Transcribed Image Text:This image depicts a multiple-choice question commonly found in educational materials, particularly in subjects like chemistry or physics, where temperature measurements are relevant. Below are four options listed as possible answers: 1. 75.9°C 2. 51.1°C 3. 58.1°C 4. 101°C Each option is preceded by a radio button, indicating that this format is likely part of a digital quiz or an online test. This type of question format is typically used to assess a student's understanding of a concept or the correct interpretation of data. There are no graphs or diagrams accompanying this question.
**Heat Absorption in a Gold Block**

**Scenario:**

- A gold block with a mass of 320 grams.
- Initial temperature: 24.8°C
- Heat absorbed: 311 joules

This scenario presents a classic example of the relationship between heat energy and temperature change in a substance. The provided data can be used to calculate the resulting temperature change using the formula:

\[ q = m \cdot c \cdot \Delta T \]

Where:
- \( q \) is the heat absorbed (311 joules)
- \( m \) is the mass of the gold block (320 grams)
- \( c \) is the specific heat capacity of gold (approximately 0.129 J/g°C)
- \( \Delta T \) is the change in temperature

This example can help students understand concepts in thermodynamics, such as specific heat, heat capacity, and temperature change.

**Note**: The diagram and graph that are typically associated with this type of problem are not included in this scenario but can be drawn to further illustrate the heat absorption process and the resulting temperature change of the gold block.
Transcribed Image Text:**Heat Absorption in a Gold Block** **Scenario:** - A gold block with a mass of 320 grams. - Initial temperature: 24.8°C - Heat absorbed: 311 joules This scenario presents a classic example of the relationship between heat energy and temperature change in a substance. The provided data can be used to calculate the resulting temperature change using the formula: \[ q = m \cdot c \cdot \Delta T \] Where: - \( q \) is the heat absorbed (311 joules) - \( m \) is the mass of the gold block (320 grams) - \( c \) is the specific heat capacity of gold (approximately 0.129 J/g°C) - \( \Delta T \) is the change in temperature This example can help students understand concepts in thermodynamics, such as specific heat, heat capacity, and temperature change. **Note**: The diagram and graph that are typically associated with this type of problem are not included in this scenario but can be drawn to further illustrate the heat absorption process and the resulting temperature change of the gold block.
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