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Question
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?
Solution 1
Heat is supplied to the system at a rate of 100 W.
∴Heat supplied, Q = 100 J/s
The system performs at a rate of 75 J/s.
∴Work done, W = 75 J/s
From the first law of thermodynamics, we have:
Q = U + W
Where,
U = Internal energy
∴U = Q – W
= 100 – 75
= 25 J/s
= 25 W
Therefore, the internal energy of the given electric heater increases at a rate of 25 W.
Solution 2
Here `triangle Q = 100 W = 100 "J/s"`
`triangle W = 75"J/s"`
Since `triangle Q = triangle uu + triangle W`
= 100 - 75 = 25 J/s
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