JEE Main 2025 · 8 April · evening · Single correct · +4 / −1

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/s², specific heat of water = 4200 J/(kg K))

  1. (a) 0.36 K
  2. (b) 0.23 K
  3. (c) 0.48 K
  4. (d) 0.14 K
Answer and solution

Answer (c) 0.48 K

Solution

Physics sir · 3 steps

All the potential energy the water loses becomes heat in that same water, so the mass cancels out.

  1. Energy balance. A mass m of water falling h loses potential energy mgh. No heat leaves the pool, so all of it warms that same mass: mgh = mc Δ T.

    Δ T = gh/c

  2. Put in numbers. Δ T = (10 × 200)/4200 = 2000/4200 = 0.476 K.
  3. So. The water warms by about 0.48 K, which is option (c).

The idea to keep. When a body's own mechanical energy turns into its own heat, write mgh = mc Δ T. The mass cancels, so Δ T = gh/c and you never need m.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Thermometry & Calorimetry

Asked about 50 times in JEE Main since 2020. This question is from 2025.