Chapter 5: Problem 16
Calculate \(q,\) and determine whether heat is absorbed or released when a system does work on the surroundings equal to \(64 \mathrm{~J}\) and \(\Delta U=213 \mathrm{~J}\).
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Internal Energy
To identify changes in internal energy, we look at how much heat is absorbed or released and how much work is done. In our exercise, the internal energy change is \( \Delta U = 213 \text{ J} \). This value tells us how much the system's total energy has increased or decreased during the process.
Internal energy is essential for understanding how systems exchange energy with their surroundings through heat and work.
Heat Absorption
In the given problem, the task is to calculate \( q \). From the First Law of Thermodynamics, we have \( q + w = \Delta U \) or rearranged \( q = \Delta U - w \). Here, \( \Delta U \) is 213 J, and work \( w \) is negative because the system is doing work on its surroundings, making \( w = -64 \text{ J} \). Thus:
- \( q = 213 \text{ J} - (-64 \text{ J}) \)
- \( q = 213 \text{ J} + 64 \text{ J} \)
- \( q = 277 \text{ J} \)
Work Done by System
In the provided exercise, the system does work equal to 64 J on the surroundings. Since the work is done by the system on the surroundings, it is considered as negative work in the formula. This is why \( w = -64 \text{ J} \).
- The negative sign in work done indicates energy leaving the system.
- Positive work would suggest energy given to the system from the surroundings.