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A system gains 657 kJ of heat, resulting in a change in internal energy of the system equal to +221 kJ. How much work is done?

A system gains 657 kJ of heat, resulting in a change in internal energy of the system equal to +221 kJ. How much work is done?

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: Work and Heat Transfer** In this exercise, you will calculate the work done on or by a system based on changes in heat and internal energy. --- **Problem Statement:** A system gains 657 kJ of heat, resulting in a change in internal energy of the system equal to +221 kJ. **Question:** How much work is done? \[ w = \_\_\_\_ \, \text{kJ} \] --- **Select the Correct Statement:** - \( \bigcirc \) Work was done by the system. - \( \bigcirc \) Work was done on the system. --- **Analysis:** Use the first law of thermodynamics, which is expressed as: \[ \Delta U = Q - W \] Where: - \( \Delta U \) is the change in internal energy. - \( Q \) is the heat added to the system. - \( W \) is the work done by the system. Given: - \( \Delta U = +221 \, \text{kJ} \) - \( Q = 657 \, \text{kJ} \) Rearrange the equation to find the work done: \[ W = Q - \Delta U \]
Transcribed Image Text:**Thermodynamics Problem: Work and Heat Transfer** In this exercise, you will calculate the work done on or by a system based on changes in heat and internal energy. --- **Problem Statement:** A system gains 657 kJ of heat, resulting in a change in internal energy of the system equal to +221 kJ. **Question:** How much work is done? \[ w = \_\_\_\_ \, \text{kJ} \] --- **Select the Correct Statement:** - \( \bigcirc \) Work was done by the system. - \( \bigcirc \) Work was done on the system. --- **Analysis:** Use the first law of thermodynamics, which is expressed as: \[ \Delta U = Q - W \] Where: - \( \Delta U \) is the change in internal energy. - \( Q \) is the heat added to the system. - \( W \) is the work done by the system. Given: - \( \Delta U = +221 \, \text{kJ} \) - \( Q = 657 \, \text{kJ} \) Rearrange the equation to find the work done: \[ W = Q - \Delta U \]
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