Question

3.2 thermodynamics at constant pressure, a piston expanded and absorbed 1000 j of heat. if δe = -1500 j, how much work was involved?
-2500 j
-500 j
2500 j
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Explanation:

Step1: Recall the first law of thermodynamics

The first law of thermodynamics is given by the equation $\Delta E = q + w$, where $\Delta E$ is the change in internal energy, $q$ is the heat added to the system, and $w$ is the work done on the system.

Step2: Determine the sign of $q$

The piston absorbed 1000 J of heat, so heat is added to the system. Therefore, $q = + 1000\ J$. We are given that $\Delta E=-1500\ J$.

Step3: Solve for $w$

We rearrange the first law equation to solve for $w$: $w=\Delta E - q$.

Substitute the known values: $w=- 1500\ J-1000\ J=-2500\ J$.

Answer:

-2500 J