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------------------------------------------------ How much work is associated with a process where 100.2 kJ of heat is absorbed and the internal energy of the system increases by 134.9 kJ?

The work associated with the process is 34.7 kJ, and it is done on the system. This is calculated using the first law of thermodynamics, using given values for the heat absorbed by the system and the increase in the system's internal energy.

The question relates to the first law of thermodynamics, which states that the internal energy of a system is equal to the heat absorbed by the system minus the work done by the system. In mathematical terms: ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat absorbed, and W is the work done. Plug the given values into the equation:

134.9 kJ = 100.2 kJ - W

Solve the equation for W (the work done):

W = 100.2 kJ - 134.9 kJ = -34.7 kJ

The negative sign suggests that work is being done on the system, not by the system. Therefore, the work associated with this process is 34.7 kJ done on the system.

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