An electric appliance supplies 6000J min ^ - 1 , heat to the system. If the system delivers a power of 90W . How long it would take to increase the internal energy by 2 . 5 × 10 ^ 3 J ?

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Correct answer: B. 2 . 5 × 10 ^ 2 s

An electric appliance supplies 6000J min ^ - 1 , heat to the system. If the system delivers a power of 90W . How long

Recall the formula of the first law of thermodynamics in terms of change in internal energy and work done by the system, $\DeltaU = Q - W$ , given, $\DeltaU = 2500J$ , $Q = \dfrac{6000}{60} \times t =$ \left(100t\right) \mid J $and$ W = P \times t = 90t $, <br />$ ⇒ 2500J = 100tJ - 90tJ ⇒ t = 2 . 5 \times {10}^{2}s.

An electric appliance supplies 6000J min ^ - 1 , heat to the...

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