Q 11-10-131JEE MainJEE Main 2025 (8 Apr, Shift 2)Easy
Water falls from a height of $200\ \text{m}$ into a pool. Calculate the rise in temperature of the water, assuming no heat dissipation from the water in the pool. (Take $g = 10\ \text{m/s}^2$, specific heat of water = $4200\ \text{J/(kg K)}$)
Answer: (D) $0.48\ \text{K}$
All the potential energy lost becomes heat in the water:
$$mgh = ms\Delta T \Rightarrow \Delta T = \frac{gh}{s} = \frac{10\times200}{4200} \approx 0.48\ \text{K}$$
Solution by Sreeraj P, M.Sc Physics