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 $$2.5 \times 10^3$$ J?

Given data:

Heat supplied to the system: $$\dot Q = 6000 \text{ J min}^{-1}$$
Useful power (work done by the system): $$P = 90 \text{ W} = 90 \text{ J s}^{-1}$$
Required rise in internal energy: $$\Delta U = 2.5 \times 10^3 \text{ J}$$

Step 1: Convert the heat-input rate to SI units (joules per second).

$$6000 \text{ J min}^{-1} = \frac{6000 \text{ J}}{60 \text{ s}} = 100 \text{ J s}^{-1}$$

Step 2: Write the First Law of Thermodynamics in its rate form.

The first law states $$\Delta Q = \Delta U + \Delta W$$. Dividing by $$\Delta t$$ gives the power form
$$\frac{\Delta Q}{\Delta t} = \frac{\Delta U}{\Delta t} + \frac{\Delta W}{\Delta t}$$
or $$\dot Q = \dot U + \dot W$$ $$-(1)$$

Step 3: Substitute the known rates into equation $$-(1)$$.

Heat input rate: $$\dot Q = 100 \text{ J s}^{-1}$$
Work output rate: $$\dot W = 90 \text{ J s}^{-1}$$
Therefore,
$$\dot U = \dot Q - \dot W = 100 - 90 = 10 \text{ J s}^{-1}$$

Step 4: Find the time needed to raise the internal energy by $$\Delta U = 2.5 \times 10^3 \text{ J}$$.

Using $$\dot U = \frac{\Delta U}{\Delta t}$$, we get
$$\Delta t = \frac{\Delta U}{\dot U} = \frac{2.5 \times 10^3}{10} = 250 \text{ s}$$

Answer: $$\boxed{2.5 \times 10^2 \text{ s}}$$   (Option B)