Question

determine the sign for q and w in the description below.
a system releases 18 kj of heat to its surroundings and the surroundings do 35 kj of work on the system.
q is
w is
calculate δe for the description of the process above.
kj

Explanation:

Step1: Determine the sign of \( q \)

When a system releases heat to the surroundings, \( q<0 \). Here, the system releases heat, so \( q=- 18\ \text{kJ}\).

Step2: Determine the sign of \( w \)

When the surroundings do work on the system, \( w>0 \). Since the surroundings do work on the system, \( w = + 35\ \text{kJ}\).

Step3: Calculate \(\Delta E\)

Use the first law of thermodynamics \(\Delta E=q + w\).
Substitute \(q=-18\ \text{kJ}\) and \(w = 35\ \text{kJ}\) into the formula:
\(\Delta E=-18\ \text{kJ}+35\ \text{kJ}\)
\(\Delta E = 17\ \text{kJ}\)

Answer:

For \(q\): negative (\(-\)), for \(w\): positive (\(+\)), \(\Delta E = 17\ \text{kJ}\)