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An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s, then the rate at which internal energy increases will be:
Detailed Solution
From the first law of thermodynamics: $\frac{dQ}{dt} = \frac{dU}{dt} + \frac{dW}{dt}$
$100 = \frac{dU}{dt} + 75$
$\frac{dU}{dt} = 100 - 75 = 25$ W