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

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Heat & Thermodynamics

The heat extracted out of x gram of water initially at

5 0^{\circ} C

to cool it down to

0^{\circ} C

is sufficient to evaporate (

1000 - x

) gram of water also initially at

5 0^{\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)

Answer: 922

Let

c = 4200 J/kg K

and

L = 2256 \times 1 0^{3} J/kg

. The heat released by

x

gram of water cooling from

5 0^{\circ} C

0^{\circ} C

x c (50)

. This must first raise

(1000 - x)

gram from

5 0^{\circ} C

10 0^{\circ} C

and then vaporise it, requiring

(1000 - x) c (50) + (1000 - x) L

. Equating and dividing by

c (50)

x = (1000 - x) (1 + \frac{L}{c ( 50 )}) = (1000 - x) (1 + \frac{2256 \times 1 0 ^{3}}{4200 \times 50}) = (1000 - x) (11.743)

. Solving,

12.743 x = 11743

, so

x = 921.5

. The value of

x

to the closest integer is

922

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Integer answer