A water cooler of storage capacity $120$ litres can cool water at a constant rate of $P$ watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly $3\ \mathrm{kW}$ of heat (thermal load). The temperature of water fed into the device cannot exceed $30^\circ\mathrm{C}$ and the entire stored $120$ litres of water is initially cooled to $10^\circ\mathrm{C}$. The entire system is thermally insulated. The minimum value of $P$ (in watts) for which the device can be operated for $3$ hours is (Specific heat of water is $4.2\ \mathrm{kJ\,kg^{-1}\,K^{-1}}$ and the density of water is $1000\ \mathrm{kg\,m^{-3}}$)

Step 1: Calculate the mass and heat capacity of the water.

The volume of water is $120\text{ L}$, which is equivalent to $0.120\text{ m}^3$.

$$m = \rho V = 1000 \times 0.120 = 120\text{ kg}$$ $$mc = 120 \times 4200 = 5.04 \times 10^{5}\text{ J/K}$$

Step 2: Determine the net heating rate of the system.

The external device adds heat at a rate of $3000\text{ W}$ and the cooler removes heat at a rate of $P\text{ W}$.

$$\text{Net heating rate} = 3000 - P$$

Step 3: Formulate the temperature of the water over time.

The temperature $T(t)$ at any time $t$ (in seconds) is determined by the net heat added to the water.

$$T(t) = 10 + \frac{(3000 - P)t}{mc}$$

Step 4: Apply the temperature limit condition.

The temperature of the water must not exceed $30^\circ\text{C}$ after $3\text{ hours}$ of operation.

$$t = 3 \times 3600 = 10800\text{ s}$$ $$T(3\text{ h}) \le 30^\circ\text{C}$$ $$10 + \frac{(3000 - P)(10800)}{mc} \le 30$$

Step 5: Solve for the minimum cooling power $P$.

The maximum allowable temperature rise is $\Delta T = 30^\circ\text{C} - 10^\circ\text{C} = 20^\circ\text{C}$.

$$mc \Delta T = 120 \times 4200 \times 20 = 1.008 \times 10^{7}\text{ J}$$ $$P_{\min} = 3000 - \frac{mc \Delta T}{10800}$$ $$P_{\min} = 3000 - \frac{1.008 \times 10^{7}}{10800}$$ $$P_{\min} \approx 2067\text{ W}$$

Answer: The minimum value of $P$ is 2067 W (or approximately 2067).