12.8 An electric heater supplies heat to a system at a rate of 100W. If the system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
It is given in the question that:
Heat supplied, Q = 100 J/s
Work done, W = 75 J/s
From the first law of thermodynamics, we have:
Q = U + W
U = Q – W
= 100 – 75
= 25 J/s
Therefore, the internal energy of the given electric heater increases at a rate of 25 J/s.
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