The heat extracted out of x gram of water initially at 50 °C to cool it down to 0 °C is sufficient to evaporate (1000 - x) gram of water also initially at 50 °C. The value of x (closest integer) is _______. (Take latent heat of water 2256 kJ/kg. K, specific heat capacity of water 4200 J/kg. K)

The correct answer is Option 922. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q47 – JEE-MAIN 2026

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Home Physics ThermodynamicsQuestion #513

The heat extracted out of

x

gram of water initially at

50^{\circ} C

to cool it down to

0^{\circ} C

is sufficient to evaporate

(1000 - x)

gram of water also initially at

50^{\circ} C

. The value of

x

(closest integer) is _______. (Take latent heat of water

2256 kJ/kg. K

, specific heat capacity of water

4200 J/kg. K

)

Enter Calculated Numerical Value:

Clue 1 of 3: Energy Conservation Principle

The heat extracted from one process is equal to the heat absorbed by another process.

🧠 Full Solution Path

Step 1: Calculate Heat Extracted (

Q_{1}

)✦ Active

The heat extracted from $x$ grams of water when cooled from $50^{\circ} C$ to $0^{\circ} C$ is given by $Q_{1} = m_{1} c Δ T_{1}$ . Convert mass to kg and use the given specific heat capacity.

m_{1} = x g = x \times 10^{- 3} kg

c = 4200 J/kg. K

Δ T_{1} = 50^{\circ} C - 0^{\circ} C = 50 K

Q_{1} = (x \times 10^{- 3}) \times 4200 \times 50 = 210 x J

Step 2: Calculate Heat Required for Evaporation (

Q_{2}

)○ Expand

The heat required to evaporate $(1000 - x)$ grams of water initially at $50^{\circ} C$ involves two parts: heating it from $50^{\circ} C$ to $100^{\circ} C$ and then evaporating it at $100^{\circ} C$ . The total heat is $Q_{2} = m_{2} c Δ T_{2} + m_{2} L_{v}$ . Convert mass to kg and latent heat to J/kg.

m_{2} = (1000 - x) g = (1000 - x) \times 10^{- 3} kg

c = 4200 J/kg. K

Δ T_{2} = 100^{\circ} C - 50^{\circ} C = 50 K

L_{v} = 2256 kJ/kg = 2256 \times 10^{3} J/kg

Q_{2} = ((1000 - x) \times 10^{- 3}) \times (4200 \times 50 + 2256 \times 10^{3})

Q_{2} = ((1000 - x) \times 10^{- 3}) \times (210000 + 2256000)

Q_{2} = ((1000 - x) \times 10^{- 3}) \times 2466000 = (1000 - x) \times 2466 J

Step 3: Equate Heat and Solve for

x

○ Expand

According to the problem statement, the heat extracted ( $Q_{1}$ ) is sufficient for the evaporation process ( $Q_{2}$ ). Equate $Q_{1}$ and $Q_{2}$ and solve for $x$ .

Q_{1} = Q_{2}

210 x = (1000 - x) \times 2466

210 x = 2466000 - 2466 x

210 x + 2466 x = 2466000

2676 x = 2466000

x = \frac{2466000}{2676} \approx 921.5246

The closest integer value for $x$ is $922$ .

💡 Teacher's Secret Hint

Ensure all units are consistent (e.g., SI units) before performing calculations to avoid errors.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 47 – JEE-MAIN 2026

Clue 1 of 3: Energy Conservation Principle

🧠 Full Solution Path