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

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Recall the formula of the first law of thermodynamics in terms of change in internal energy and work done by the system, $Δ U = Q - W$ , given, $Δ U = 2500 J$ , $Q = \frac{6000}{60} \times t = (100 t) J$ and $W = P \times t = 90 t$ ,
$\Rightarrow 2500 J = 100 t J - 90 t J \Rightarrow t = 2.5 \times 10^{2} s$ .

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