JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Heat & Thermodynamics

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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)

)

(A)0.23 K
(B)0.36 K
(C)0.14 K
(D)0.48 K✓ Correct

Answer: (D)

Solution

All the gravitational potential energy lost in the fall appears as heat in the water, so

m g h = m s Δ T

with

s

the specific heat. The mass cancels, leaving

Δ T = \frac{g h}{s} = \frac{10 \times 200}{4200} = \frac{10}{21}

. The rise in temperature of the water is about

0.48 K

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option

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