Question

1. what is the change in internal energy, \\( \delta u \\), for a system that absorbs 127j of heat from the surroundings and does 325j of work on the surroundings?

a. +452 j
b. +198 j
c. -198 j
d. -452 j
e. +213 j

Explanation:

Step1: Determine the sign of heat and work

Heat absorbed by the system, \( q = + 127\,\text{J}\) (positive as heat is added to the system). Work done by the system, \( w=-325\,\text{J}\) (negative as work is done by the system on the surroundings).

Step2: Use the first law of thermodynamics

The first law of thermodynamics is \(\Delta U=q + w\). Substitute \(q = 127\,\text{J}\) and \(w=-325\,\text{J}\) into the formula: \(\Delta U=127+(- 325)\).

Step3: Calculate the value of \(\Delta U\)

\(\Delta U=127 - 325=-198\,\text{J}\)

Answer:

C. -198 J