A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is $700 W m^{-2}$ and it is absorbed by the water over an effective area of $0.05 m^{2}.$ Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in °C) in the temperature of water and the surroundings after a long time will be ______. (Ignore effect of the container, and take constant for Newton's law of cooling $= 0.001 s^{-1},$ Heat capacity of water $= 4200 J$ kg$^{-1} K^{-1})$

Step 1: Identify the given parameters.

The parameters for the water and the environment are given as follows.

$$m = 1\,\text{kg}$$ $$I = 700\,\text{W m}^{-2}$$ $$A = 0.05\,\text{m}^{2}$$ $$c = 4200\,\text{J kg}^{-1}\text{K}^{-1}$$ $$k = 0.001\,\text{s}^{-1}$$

Step 2: Calculate the rate of heat energy absorbed.

The power received from sunlight is the product of solar intensity and the effective area.

$$P_{\text{in}} = I \times A$$ $$P_{\text{in}} = 700 \times 0.05$$ $$P_{\text{in}} = 35\,\text{W}$$

Step 3: Formulate the energy balance equation.

The rate of temperature change is determined by the net heat transfer rate.

$$mc\frac{dT}{dt} = P_{\text{in}} - mck(T - T_s)$$

Step 4: Apply the steady-state condition.

After a long time, the temperature of the water reaches a constant value.

$$\frac{dT}{dt} = 0$$

This simplifies the energy balance equation.

$$P_{\text{in}} = mck\Delta T$$ $$\Delta T = \frac{P_{\text{in}}}{mck}$$

Step 5: Calculate the temperature difference.

Substitute the given values into the steady-state equation.

$$\Delta T = \frac{35}{1 \times 4200 \times 0.001}$$ $$\Delta T = \frac{35}{4.2}$$ $$\Delta T = \frac{25}{3} \approx 8.33\,^{\circ}\text{C}$$

Answer: $8.33\,^{\circ}\text{C}$ (or $\frac{25}{3}\,^{\circ}\text{C}$)