Question

For a particular process, q = -10 kJ and w = 25 kJ. Which of the following statements is true?

A. ΔE = -35 kJ
B. System flows from the surroundings to the heat
C. The surroundings do work on the system
D. ΔE = -15 kJ
E. The system does work on the surroundings

Answer 1

The change in internal energy for a process with q = -10 kJ and w = 25 kJ is ΔE = 15 kJ, indicating that the surroundings do work on the system, and the system releases heat to the surroundings.

Step-by-step explanation:

Understanding the First Law of Thermodynamics

For the given process where q = -10 kJ (heat is released by the system) and w = 25 kJ (work is done on the system), we can use the first law of thermodynamics to find the change in internal energy. According to the first law, which is given by the equation ΔE = q + w, the change in internal energy (ΔE) of a system is equal to the heat added to the system (q) plus the work done on the system (w).

In this case, the correct statement is that the surroundings do work on the system since w is positive. Additionally, since q is negative, it indicates that heat is transferred from the system to the surroundings. The change in internal energy of the system is thus ΔE = -10 kJ (q) + 25 kJ (w) = 15 kJ, meaning the internal energy increases. Therefore, the correct statement is D. ΔE = 15 kJ.