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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
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.
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.
**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
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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