Question

In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Answer 1

Given:

• Heat absorbed by system: \(q = +701\ \text{J}\) (positive, as system gains heat)
• Work done by system: \(w = +394\ \text{J}\) (positive, work done by system on surroundings)

First Law of Thermodynamics:

\[ \Delta U = q - w \] (Here, \(w\) is work done by the system.)

Calculation:

\[ \Delta U = 701\ \text{J} - 394\ \text{J} = 307\ \text{J} \]

The change in internal energy for the process is: \[ \boxed{\Delta U = +307\ \text{J}} \]