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 \(2256\text{ kJ/kg.K}\), specific heat capacity of water \(4200\text{ J/kg.K}\))

Heat extracted to cool \(x\) gram of water from \(50^{\circ}\text{C}\) to \(0^{\circ}\text{C}\) is given by:

\(Q_1 = m_1 c \Delta T_1\)

\(Q_1 = x \times 4.2 \times (50 - 0) = 210x \text{ J}\)

Heat required to raise the temperature of \((1000 - x)\) gram of water from \(50^{\circ}\text{C}\) to \(100^{\circ}\text{C}\) and evaporate it is given by:

\(Q_2 = m_2 c \Delta T_2 + m_2 L\)

\(Q_2 = (1000 - x) \times [4.2 \times (100 - 50) + 2256]\)

\(Q_2 = (1000 - x) \times (210 + 2256) = 2466(1000 - x) \text{ J}\)

Equating the heat extracted and the heat required:

\(Q_1 = Q_2\)

\(210x = 2466(1000 - x)\)

\(210x = 2466000 - 2466x\)

\(2676x = 2466000\)

\(x = \dfrac{2466000}{2676} \approx 921.52\)

The closest integer value for \(x\) is \(922\).

Answer: \(922\)