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If a system has a change in energy of 400 J and produces 200 J of work, how much heat was added to the system? 800 J 200 J -200 J 600 J

If a system has a change in energy of 400 J and produces 200 J of work, how much heat was added to the system? 800 J 200 J -200 J 600 J

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Educational Content: Energy and Work Calculation** --- **Problem Statement:** If a system has a change in energy of 400 J and produces 200 J of work, how much heat was added to the system? **Options:** - 800 J - 200 J - -200 J - 600 J **Answer Explanation:** To determine how much heat was added to the system, we can use the first law of thermodynamics which can be stated as: \[ \Delta U = Q - W \] where: - \(\Delta U\) is the change in internal energy of the system. - \(Q\) is the heat added to the system. - \(W\) is the work done by the system. Given: - \(\Delta U = 400 \text{ J}\) - \(W = 200 \text{ J}\) We need to find \(Q\). Rearranging the equation for \(Q\): \[ Q = \Delta U + W \] Substituting the given values: \[ Q = 400 \text{ J} + 200 \text{ J} \] \[ Q = 600 \text{ J} \] Therefore, the correct answer is: \[ \boxed{600 \text{ J}} \] --- By understanding this calculation, students can grasp the fundamental principles of the first law of thermodynamics and the relationship between energy, heat, and work within a system.
Transcribed Image Text:**Educational Content: Energy and Work Calculation** --- **Problem Statement:** If a system has a change in energy of 400 J and produces 200 J of work, how much heat was added to the system? **Options:** - 800 J - 200 J - -200 J - 600 J **Answer Explanation:** To determine how much heat was added to the system, we can use the first law of thermodynamics which can be stated as: \[ \Delta U = Q - W \] where: - \(\Delta U\) is the change in internal energy of the system. - \(Q\) is the heat added to the system. - \(W\) is the work done by the system. Given: - \(\Delta U = 400 \text{ J}\) - \(W = 200 \text{ J}\) We need to find \(Q\). Rearranging the equation for \(Q\): \[ Q = \Delta U + W \] Substituting the given values: \[ Q = 400 \text{ J} + 200 \text{ J} \] \[ Q = 600 \text{ J} \] Therefore, the correct answer is: \[ \boxed{600 \text{ J}} \] --- By understanding this calculation, students can grasp the fundamental principles of the first law of thermodynamics and the relationship between energy, heat, and work within a system.
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