The heat extracted out of x gram of water initially at 50°C to cool...

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: $922$