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Answer :
To solve this problem, we need to understand the process of heat exchange between steam and water. When steam condenses to water, it releases latent heat, which is then used to raise the temperature of the initial water.
Initial Data:
- Mass of initial water: [tex]m_1 = 22 \text{ g}[/tex]
- Initial temperature of water: [tex]T_1 = 20^\circ\text{C}[/tex]
- Final temperature of water: [tex]T_2 = 90^\circ\text{C}[/tex]
- Latent heat of steam ([tex]L[/tex]): [tex]540 \text{ cal/g}[/tex]
- Specific heat of water ([tex]s[/tex]): [tex]1 \text{ cal/g}^\circ\text{C}[/tex]
Calculating Heat Required to Raise Temperature:
The heat needed to increase the initial water from [tex]20^\circ\text{C}[/tex] to [tex]90^\circ\text{C}[/tex] can be calculated using the formula:
[tex]Q_1 = m_1 \cdot s \cdot (T_2 - T_1)[/tex]
[tex]Q_1 = 22 \text{ g} \cdot 1 \text{ cal/g}^\circ\text{C} \cdot (90 - 20)^\circ\text{C} = 1540 \text{ cal}[/tex]
Calculating Steam Needed to Condense:
The heat released by steam when it condenses to water and then cools down to [tex]90^\circ\text{C}[/tex] is given by:
[tex]Q_2 = m_2 \cdot L[/tex]
where [tex]m_2[/tex] is the mass of the steam. Since the converted steam will also come down to [tex]90^\circ\text{C}[/tex], we don’t need to consider additional heat loss here as it is already calculated in [tex]Q_1[/tex].
Calculate [tex]m_2[/tex]:
The heat gained by water must be equal to the heat lost by condensing steam:
[tex]Q_1 = Q_2 \\
1540 \text{ cal} = m_2 \cdot 540 \text{ cal/g}[/tex]Solving for [tex]m_2[/tex]:
[tex]m_2 = \frac{1540}{540} \approx 2.85 \text{ g}[/tex]
Total Mass of Water at 90°C:
The total mass of water when the mixture reaches [tex]90^\circ\text{C}[/tex] is the initial mass of water plus the mass of the condensed steam:
[tex]m_{total} = m_1 + m_2 = 22 \text{ g} + 2.85 \text{ g} \approx 24.85 \text{ g}[/tex]
Since the closest option to our calculated result is [tex]24.8 \text{ g}[/tex], the answer is option (1) 24.8 g.
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