Calculating Heat Transfer for Cooling and Evaporation

The problem requires calculating the mass ($x$) of water based on heat transfer principles. Two main processes are involved:

• Cooling a mass of water ($x$ grams) from $50 \ ^{\circ}\text{C}$ down to $0 \ ^{\circ}\text{C}$.
• Evaporating another mass of water ($(1000 - x)$ grams) initially at $50 \ ^{\circ}\text{C}$.

The core principle is that the heat energy extracted during the cooling process is equal to the heat energy required for the evaporation process.

Heat Extracted During Cooling

The heat ($Q_{cool}$) removed from water when its temperature changes is given by the formula:

$Q_{cool} = m \cdot c \cdot \Delta T$

Using the provided values:

• Mass ($m$) = $x$ grams $= \frac{x}{1000}$ kg
• Specific heat capacity of water ($c$) = $4200 \text{ J/kg. K}$
• Temperature change ($\Delta T$) = $50 \ ^{\circ}\text{C} - 0 \ ^{\circ}\text{C} = 50 \ ^{\circ}\text{C}$ (which is equal to $50 \text{ K}$)

Substitute these into the formula:

$Q_{cool} = \left(\frac{x}{1000} \text{ kg}\right) \times (4200 \text{ J/kg. K}) \times (50 \text{ K})$

Simplify the calculation:

$Q_{cool} = \frac{x \times 4200 \times 50}{1000} \text{ J} = 210x \text{ J}$

Heat Required for Evaporation

The heat ($Q_{evap}$) needed to change the state of water from liquid to gas (evaporation) at a constant temperature is given by:

$Q_{evap} = m_{evap} \cdot L_v$

Using the provided values:

• Mass ($m_{evap}$) = $(1000 - x)$ grams $= \frac{1000 - x}{1000}$ kg
• Latent heat of vaporization ($L_v$) = $2256 \text{ kJ/kg}$. (Note: The unit '. K' in the question is assumed to be a typo for latent heat). Convert kJ/kg to J/kg: $L_v = 2256 \times 1000 \text{ J/kg}$.

Substitute these into the formula:

$Q_{evap} = \left(\frac{1000 - x}{1000} \text{ kg}\right) \times (2256 \times 1000 \text{ J/kg})$

Simplify the calculation:

$Q_{evap} = (1000 - x) \times 2256 \text{ J}$

Equating Heat Transfer

As stated, the heat extracted equals the heat required:

$Q_{cool} = Q_{evap}$

Substitute the expressions for $Q_{cool}$ and $Q_{evap}$:

$210x = (1000 - x) \times 2256$

Solving for x

Now, solve the algebraic equation for $x$:

$210x = 2256 \times 1000 - 2256x$

$210x = 2256000 - 2256x$

Combine terms involving $x$:

$210x + 2256x = 2256000$

$2466x = 2256000$

Isolate $x$:

$x = \frac{2256000}{2466}$

Calculate the value of $x$:

$x \approx 914.84$

Final Answer Determination

The question asks for the value of $x$ as the closest integer. Rounding the calculated value $914.84$ to the nearest integer yields $915$.