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If you do 1,000 J of work on a system and remove 500 J of heat from it, what is the change in its internal energy? -1500 J, -500 J, 500 J, or 1500 J thank you!!
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here we have to apply the first principle of thermodynamics: \[\Large Q = L + \Delta U\] where \Delta U is the internal energy change
okay!! so 1000+500? so the solution is 1500 J ?
more explanation: since do the work L on the thermodynamic system, then L=-1000 J, namely it is negative, so we have: 500=-1000+ \Delta U then: \Delta U = 1500 Joules so your answer is correct!
oops.. since we do the work on the thermodynamic system...
oooh yay!! :D thank you:)
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thank you!! :)
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