Question

what is the change in internal energy, \\( \delta u \\), in joules, for a system that absorbs 575 j of heat and does 415 j of work on its surroundings?

• 160 j

+ 575 j

• 415 j

+ 990 j
+ 160 j

Explanation:

Step1: Determine the sign of heat and work

Heat absorbed by the system, \( q = + 575\,\text{J}\) (positive as heat is added to the system).
Work done by the system on the surroundings, \( w=-415\,\text{J}\) (negative as work is done by the system).

Step2: Use the first law of thermodynamics

The first law of thermodynamics is \(\Delta U=q + w\).
Substitute \(q = 575\,\text{J}\) and \(w=-415\,\text{J}\) into the formula:
\(\Delta U=575+(- 415)\)
\(\Delta U=575 - 415\)
\(\Delta U = 160\,\text{J}\)

Answer:

+ 160 J