Physics Q32 – JEE-MAIN 2025

Home Physics ThermodynamicsQuestion #536

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)

)

Clue 1 of 3: Energy Transformation

Consider how the potential energy of the falling water is converted into another form of energy upon impact.

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Step 1: Identify Energy Transformation✦ Active

When water falls from a height, its potential energy is converted into kinetic energy, and upon impact with the pool, this kinetic energy is dissipated as heat, increasing the water's internal energy. Assuming no heat dissipation to the surroundings, all the potential energy lost by the water is converted into its thermal energy.

P E_{l o s t} = Q_{g a i n e d}

Step 2: Formulate Energy Equations○ Expand

The potential energy lost by a mass $m$ of water falling from height $h$ is $P E = m g h$ . The heat gained by the same mass $m$ of water, causing a temperature rise of $Δ T$ , is $Q = m c Δ T$ , where $c$ is the specific heat capacity of water.

m g h = m c Δ T

Step 3: Calculate Temperature Rise○ Expand

From the energy conservation equation, we can cancel out the mass $m$ and solve for $Δ T$ :

Δ T = \frac{g h}{c}

Substitute the given values: $g = 10 {m/s}^{2}$ , $h = 200 m$ , and $c = 4200 J/(kg K)$ .

Δ T = \frac{10 \times 200}{4200} = \frac{2000}{4200} = \frac{20}{42} = \frac{10}{21} \approx 0.476 K

Rounding to two decimal places, the rise in temperature is approximately $0.48 K$ .

💡 Teacher's Secret Hint

Ensure units are consistent throughout the calculation to avoid errors.

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Physics Question 32 – JEE-MAIN 2025

Clue 1 of 3: Energy Transformation

🧠 Full Solution Path