What is the difference in temperature of water at the top and bottom of a rainfall of height 500 mts.

(Sp. Heat of water = $$10^3$$​ cal/kg & J = 4.2 J/cal)

Correct option is C

​$$\textbf{Given:} \\\bullet \text{Height, } h = 500\, \text{m} \\\bullet \text{Specific heat, } C = 10^3 \times 4.2 = 4200\, \text{J/kg}^\circ \text{C} \\\bullet \text{Acceleration due to gravity, } g = 9.81\, \text{m/s}^2\\[10pt]\text{By conservation of energy:} \\[5pt]\text{Loss in potential energy} = \text{Gain in internal energy} \\[5pt]mgh = mC \Delta T\\[10pt]\text{Cancel mass } m \text{ from both sides:} \\[5pt]gh = C \Delta T \Rightarrow \Delta T = \frac{gh}{C} = \frac{9.81 \times 500}{4200} = \frac{4905}{4200} \approx 1.168^\circ \text{C}$$​​