Question:
The temperature of water of mass 100 g is raised from 24°C to 90°C by adding steam to it. The mass of the steam added is (Latent heat of steam = 540 cal/g and specific heat capacity of water = 1 cal/g°C)
The temperature of water of mass 100 g is raised from 24°C to 90°C by adding steam to it. The mass of the steam added is (Latent heat of steam = 540 cal/g and specific heat capacity of water = 1 cal/g°C)
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Heat lost = heat gained. Heat lost by steam: $mL_v + mc\Delta T$. Heat gained by water: $Mc\Delta T$.
Updated On: Jun 5, 2025
- 10 g
- 12 g
- 8 g
- 16 g
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The Correct Option is B
Solution and Explanation
Let $m$ be the mass of steam added. Heat lost by steam = Heat gained by water. Heat lost by steam = $mL_v + mc(100-90)$, where $L_v$ is the latent heat of vaporization and $c$ is the specific heat of water. Heat gained by water = $Mc(90-24)$, where $M$ is the mass of water. $m(540) + m(1)(10) = (100)(1)(66)$ $550m = 6600$ $m = \frac{6600}{550} = 12$ g.
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