The heat extracted out of x gram of water initialy at  $$50°$$C to  cool it down to $$0°$$C is sufficient to evaporate  $$(1000 - x)$$ gram  of water also initialy at $$50°$$C. The value of $$x$$ (closest integer) is_______.  (Take Latent heat of water  $$L = 2256$$ kJ/kg K, specific heat capacity of water $$c = 4200$$ J/kg·K)

$$\text{Heat Lost} = \text{Heat Gained}$$

$$Q = mc\Delta T$$ (temperature change without phase change)

$$Q = mL$$ (Phase change)

The first part of the water ($$x$$ grams) is cooled from $$50^\circ\text{C}$$ to $$0^\circ\text{C}$$.

$$Q_{\text{extracted}} = m_1 \cdot c \cdot \Delta T$$

$$Q_{\text{extracted}} = \left(x \times 10^{-3}\right) \times 4200 \times 50$$

$$Q_{\text{extracted}} = x \times 10^{-3} \times 210,000 = 210x\text{ J}$$

The remaining water, $$(1000-x)$$ grams, is initially at $$50^\circ\text{C}$$. To evaporate it, it must first be heated to its boiling point ($$100^\circ\text{C}$$) and then converted into steam at $$100^\circ\text{C}$$.

The total heat required ($$Q_{\text{required}}$$) consists of two parts:

Heating from $$50^\circ\text{C}$$ to $$100^\circ\text{C}$$ ($$\Delta T_2 = 100 - 50 = 50\text{ K}$$): $$Q_{\text{heating}} = m_2 \cdot c \cdot \Delta T_2$$

Evaporating at $$100^\circ\text{C}$$: $$Q_{\text{phase\_change}} = m_2 \cdot L$$

$$Q_{\text{required}} = m_2 \left(c\Delta T_2 + L\right)$$

$$Q_{\text{required}} = (1000 - x) \times 10^{-3} \times \left[(4200 \times 50) + \left(2256 \times 10^3\right)\right]$$

$$Q_{\text{required}} = 2466(1000 - x)\text{ J}$$

$$Q_{\text{extracted}} = Q_{\text{required}}$$

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

$$x = \frac{2,466,000}{2676}$$

$$x \approx 921.52 \approx 922$$