An electric appliance supplies 6000 J / min 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 x 10³ J?

Given, heat supplied to the system $\frac{\mathrm{\Delta Q}}{\mathrm{\Delta t}}=6000\mathrm{J}/min=\frac{6000}{60}\mathrm{J}/\mathrm{S}=100\mathrm{J}/\mathrm{S}$

Power delivered, $\mathrm{P}=\frac{\mathrm{\Delta W}}{\mathrm{\Delta t}}=90\mathrm{W}$ increase in internal energy $\mathrm{\Delta U}=2.5\times {10}^3\mathrm{J}$ 

From first law of thermodynamics, we have  $\mathrm{\Delta Q}=\mathrm{\Delta U}+\mathrm{\Delta W}\text{ Or }\frac{\mathrm{\Delta Q}}{\mathrm{\Delta t}}=\frac{\mathrm{\Delta U}}{\mathrm{\Delta t}}+\frac{\mathrm{\Delta W}}{\mathrm{\Delta t}}$ 
Substituting the given values in Eq.(i), we get $100=\frac{2.5\times {10}^3}{\mathrm{\Delta t}}+90$

$10=\frac{2.5\times {10}^3}{\mathrm{\Delta t}}$

$\mathrm{\Delta t}=\frac{2.5\times {10}^3}{10}\mathrm{\Delta t}=2.5\times {10}^2\mathrm{S}$