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MCQEasyJEE 2025Calorimetry & Change of State

JEE Physics 2025 Question with Solution

Water falls from a height of 200 m200 \, \text{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/s2g = 10 \, \text{m/s}^2, specific heat of water = 4200 J/(kg K)4200 \, \text{J/(kg K)})

  • A

    0.48 K0.48 \, \text{K}

  • B

    0.36 K0.36 \, \text{K}

  • C

    0.14 K0.14 \, \text{K}

  • D

    0.23 K0.23 \, \text{K}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Water falls through a height of 200 m200 \, \text{m}. Acceleration due to gravity is g=10 m/s2g = 10 \, \text{m/s}^2 and specific heat capacity of water is c=4200 J/(kg K)c = 4200 \, \text{J/(kg K)}.

Find: Rise in temperature of the water.

The potential energy lost by the falling water is converted entirely into thermal energy.

mgh=mcΔTmgh = mc\Delta T

Cancel mm from both sides:

gh=cΔTgh = c\Delta T

So,

ΔT=ghc\Delta T = \frac{gh}{c}

Substitute the given values:

ΔT=10×2004200\Delta T = \frac{10 \times 200}{4200}ΔT=20004200=0.476 K\Delta T = \frac{2000}{4200} = 0.476 \, \text{K}

Rounding to two decimal places,

ΔT≈0.48 K\Delta T \approx 0.48 \, \text{K}

Therefore, the rise in temperature of the water is 0.48 K0.48 \, \text{K}. The correct option is A.

Energy per Unit Mass Approach

Given: Take 1 kg1 \, \text{kg} of water falling from a height of 200 m200 \, \text{m}.

Find: Temperature rise produced by conversion of gravitational potential energy into heat.

For 1 kg1 \, \text{kg} of water, the potential energy lost is

PE=mgh=1×10×200=2000 JPE = mgh = 1 \times 10 \times 200 = 2000 \, \text{J}

This energy appears as heat gained by the water:

Q=mcΔTQ = mc\Delta T

Using m=1 kgm = 1 \, \text{kg},

2000=1×4200×ΔT2000 = 1 \times 4200 \times \Delta TΔT=20004200≈0.476 K\Delta T = \frac{2000}{4200} \approx 0.476 \, \text{K}

Hence, the required rise in temperature is approximately 0.48 K0.48 \, \text{K}.

Common mistakes

  • Using kinetic energy at impact separately and not converting it into heat is incorrect here, because the question assumes no heat dissipation and complete conversion of lost potential energy into thermal energy. Directly use mgh=mcΔTmgh = mc\Delta T.

  • Forgetting to cancel the mass mm from both sides can make the calculation look dependent on the amount of water. The temperature rise is independent of mass here, so simplify to gh=cΔTgh = c\Delta T first.

  • Using an incorrect specific heat unit or omitting units for cc can lead to dimensional inconsistency. Use 4200 J/(kg K)4200 \, \text{J/(kg K)} so that the resulting temperature change comes in K\text{K}.

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