250 kJ of heat is added to a system. The system then applied 125 kJ of work to the surroundings. How much was the change in the system's internal energy after this process?
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ΔU = Q - W
where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.
In this case, Q = 250 kJ and W = -125 kJ (since the work is done by the system on the surroundings, the sign is negative). Substituting these values into the equation, we get:
ΔU = 250 kJ - (-125 kJ)
ΔU = 375 kJ
Therefore, the change in the system's internal energy after the process is 375 kJ.