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))

Heat and Thermodynamics PYQ — JEE Main 2025 · 8 Apr · Shift 2 · Q60 with Solution

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Correct answer: C

When water falls through height $h$ , its loss in gravitational potential energy per unit mass is

$E = gh$

If all of this energy converts into heat in the water, then

$mc\Delta T = mgh$

Cancelling $m$ :

$c\Delta T = gh$

So,

$\Delta T = \frac{gh}{c}$

Given:

Thus,

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

$\Delta T = \frac{2000}{4200}$

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

$0.48\ \text{K}$ corresponds to Option C.

Stored correct answer: C

My derived answer: C

So, the derived answer agrees with the stored answer.