Question

a certain process released 25.5 kj of heat to the surroundings and did 500 j of work on the surroundings. what is the $delta e$ for the process in kilojoules?
525.5 kj
-525.5 kj
-26.0 kj
26.0 kj
-25.0 kj
25.0 kj

Explanation:

Step1: Convert work to kJ

The work done on the surroundings is $500\ J$. Since $1\ kJ = 1000\ J$, then $500\ J=\frac{500}{1000}=0.5\ kJ$.

Step2: Determine the sign of heat and work

Heat released to the surroundings means $q=- 25.5\ kJ$ (negative as heat is lost by the system). Work done on the surroundings means $w = - 0.5\ kJ$ (negative as work is done by the system).

Step3: Use the first - law of thermodynamics formula

The first - law of thermodynamics is $\Delta E=q + w$. Substitute $q=-25.5\ kJ$ and $w=-0.5\ kJ$ into the formula: $\Delta E=-25.5+( - 0.5)=-26.0\ kJ$.

Answer:

-26.0 kJ