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 $2.5\times {10}^3 \mathrm{J}$?

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

$\begin{array}{l} \mathrm{\Delta Q}=\mathrm{\Delta U}+\mathrm{\Delta W}\\ \mathrm{or} \frac{\mathrm{\Delta Q}}{\mathrm{\Delta t}}=\frac{\mathrm{\Delta U}}{\mathrm{\Delta t}}+\frac{\mathrm{\Delta W}}{\mathrm{\Delta t}} ...\left(\mathrm{i}\right)\end{array}$

Substituting the given values in Eq. (i), we get

$\begin{array}{l} 100=\frac{2.5\times {10}^3}{\mathrm{\Delta t}}+90\\ \Rightarrow 10=\frac{2.5\times {10}^3}{\mathrm{\Delta t}}\\ \Rightarrow \mathrm{\Delta t}=\frac{2.5\times {10}^3}{10}\Rightarrow \mathrm{\Delta t}=2.5\times {10}^2\mathrm{s}\end{array}$