A chemical reaction is run in which 295 Joules of work is done on the system and the internal energy changes by + 1030 Joules. Calculate q for the system. q= Joules
A chemical reaction is run in which 295 Joules of work is done on the system and the internal energy changes by + 1030 Joules. Calculate q for the system. q= Joules
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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Thermodynamics Problem: Calculating Heat Exchange (q)
**Problem Statement:**
A chemical reaction is run in which 295 Joules of work is done on the system and the internal energy changes by +1030 Joules.
**Task:**
Calculate \( q \) for the system.
\[ q = \_\_\_\_\_\_\_\_ \text{ Joules} \]
### Explanation:
To solve this problem, use the first law of thermodynamics, which is given by:
\[ \Delta U = q + w \]
Where:
- \( \Delta U \) is the change in internal energy of the system.
- \( q \) is the heat exchange in the system.
- \( w \) is the work done on or by the system.
From the problem statement:
- \( \Delta U = +1030 \) Joules
- \( w = +295 \) Joules (since work is done *on* the system, it's positive according to the sign convention).
Rearranging the equation to solve for \( q \):
\[ q = \Delta U - w \]
Substitute the given values into the equation:
\[ q = 1030 \text{ J} - 295 \text{ J} \]
Simplifying this:
\[ q = 735 \text{ J} \]
Therefore:
\[ q = 735 \text{ Joules} \]
You can input this value into the blank provided in the original problem statement.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F216d8975-823a-443e-8abc-06f056adfa09%2Fdc237ea0-a659-47d3-aef6-afe01b8e33f6%2Fv8klfs_processed.png&w=3840&q=75)
Transcribed Image Text:### Thermodynamics Problem: Calculating Heat Exchange (q)
**Problem Statement:**
A chemical reaction is run in which 295 Joules of work is done on the system and the internal energy changes by +1030 Joules.
**Task:**
Calculate \( q \) for the system.
\[ q = \_\_\_\_\_\_\_\_ \text{ Joules} \]
### Explanation:
To solve this problem, use the first law of thermodynamics, which is given by:
\[ \Delta U = q + w \]
Where:
- \( \Delta U \) is the change in internal energy of the system.
- \( q \) is the heat exchange in the system.
- \( w \) is the work done on or by the system.
From the problem statement:
- \( \Delta U = +1030 \) Joules
- \( w = +295 \) Joules (since work is done *on* the system, it's positive according to the sign convention).
Rearranging the equation to solve for \( q \):
\[ q = \Delta U - w \]
Substitute the given values into the equation:
\[ q = 1030 \text{ J} - 295 \text{ J} \]
Simplifying this:
\[ q = 735 \text{ J} \]
Therefore:
\[ q = 735 \text{ Joules} \]
You can input this value into the blank provided in the original problem statement.
Transcribed Image Text:**Question:**
A sample of solid iron is heated with an electrical coil. If 77.7 Joules of energy are added to an 11.9 gram sample and the final temperature is 38.3°C, what is the initial temperature of the iron?
**Answer:**
\( \_\_\_\_\_\_\_\) °C
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