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Q12.8 An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?

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Rate at which heat is supplied \Delta Q=100\ W

Rate at which work is done \Delta W=75\ J s^{-1}

Rate of change of internal energy is \\\Delta u

\\\Delta u=\Delta Q-\Delta w\\ \Delta u=100-75\\ \Delta u=25\ J\ s^{-1}

The internal energy of the system is increasing at a rate of 25 J s-1

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