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A system gives out 30 J of heat and also...

A system gives out 30 J of heat and also does 40 J of work. What is the internal energy change?

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To solve the problem of finding the change in internal energy (ΔU) of a system that gives out 30 J of heat and does 40 J of work, we can use the first law of thermodynamics. The first law states that: \[ \Delta U = Q + W \] Where: - \( \Delta U \) = change in internal energy - \( Q \) = heat added to the system (positive if heat is absorbed, negative if heat is released) - \( W \) = work done on the system (positive if work is done on the system, negative if work is done by the system) ...
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