The internal energy ∆E of a system is -230 kJ. If the system released 300 kJ, How much work was done in the system? Based on your answer, one of the following answer is correct.
The system does work
Heat flow is constant
Work is done on the system
It is a process were heat flow from the surrounding to the system
We can use the first law of thermodynamics to relate the energy change of a system to the heat and work involved: ∆E = Q - W, where ∆E is the internal energy change, Q is the heat added to the system, and W is the work done by the system. In this case, we have ∆E = -230 kJ and Q = -300 kJ (negative because heat is released from the system). Therefore, we can solve for W:
∆E = Q - W
-230 kJ = -300 kJ - W
W = -300 kJ + 230 kJ
W = -70 kJ
So, 70 kJ of work was done on the system (negative because the work is done by the system). Based on this answer, we can say that "work is done on the system."
To find out how much work was done in the system, we can use the first law of thermodynamics, which states that the change in internal energy (\( \Delta E \)) of a system is equal to the heat added to the system (\( Q \)) minus the work done by the system (\( W \)). Mathematically, this can be written as:
\( \Delta E = Q - W \)
Given that \( \Delta E = -230 \) kJ and \( Q = -300 \) kJ (since the system released 300 kJ), we can substitute these values into the equation to solve for \( W \):
\( -230 = -300 - W \)
Rearranging the equation, we have:
\( W = -300 + 230 = -70 \) kJ
Since the work done (\( W \)) is negative, it indicates that work was done on the system (as opposed to the system doing work). Therefore, the correct answer is:
Work is done on the system.