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An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 Joules per second. At what rate is the internal energy increasing? |
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Answer» Given ∆Q = 100w = 100J/s ∆W = 75 J/s ∆Q = ∆U + ∆W ∆U = ∆Q – ∆W = 100 – 75 = 25J/S |
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