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2897 j of heat is added to a system and 2450 j of work is done by the system. what is the change in internal energy of the system?

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SOLUTION:

∆E = Q + W

∆E = 2897 J + (-2450 J)

∆E = 2897 J - 2450 J

\boxed{\Delta E = \text{447 J}}

ΔE=447 J

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