Water falls from a height of 200 m into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. (Take $$g = 10$$ m/s$$^2$$, specific heat of water = 4200 J/(kg K))

When water falls freely, its gravitational potential energy is converted into internal energy (heat) on reaching the pool.
Ignoring any heat loss to the surroundings, the gain in internal energy raises the water’s temperature.

For a mass $$m$$ of water falling through height $$h$$, the potential energy lost is
$$\text{Potential energy} = m g h$$

This energy appears as heat: $$m c \Delta T$$, where $$c$$ is the specific heat capacity and $$\Delta T$$ is the rise in temperature.

Equating the two energies:
$$m g h = m c \Delta T$$

Cancel the common factor $$m$$ (mass of water):
$$g h = c \Delta T$$

Hence the temperature rise is
$$\Delta T = \frac{g h}{c}$$

Insert the given values: $$g = 10 \text{ m s}^{-2}$$, $$h = 200 \text{ m}$$, $$c = 4200 \text{ J kg}^{-1} \text{ K}^{-1}$$.

$$\Delta T = \frac{10 \times 200}{4200}$$

$$\Delta T = \frac{2000}{4200} = 0.476 \text{ K} \approx 0.48 \text{ K}$$

Therefore, the rise in temperature of the water is about $$0.48 \text{ K}$$.

Correct option: Option D (0.48 K)