6 gm of dteam at 100°C is mixed with 6 gm of ice at 0°C the mass of steam gets uncondensed is
Question
6 gm of dteam at 100°C is mixed with 6 gm of ice at 0°C the mass of steam gets uncondensed is
Solution
To solve this problem, we need to consider the heat transfer that occurs when the steam and ice are mixed.
Step 1: Calculate the heat absorbed by the ice to reach its melting point. The heat absorbed by the ice can be calculated using the formula: Q = m * Lf where Q is the heat absorbed, m is the mass of the ice, and Lf is the latent heat of fusion of ice.
Given: m = 6 gm (mass of ice) Lf = 334 J/gm (latent heat of fusion of ice)
Q = 6 gm * 334 J/gm Q = 2004 J
Step 2: Calculate the heat released by the steam to reach its condensation point. The heat released by the steam can be calculated using the formula: Q = m * Lv where Q is the heat released, m is the mass of the steam, and Lv is the latent heat of vaporization of steam.
Given: m = 6 gm (mass of steam) Lv = 2260 J/gm (latent heat of vaporization of steam)
Q = 6 gm * 2260 J/gm Q = 13560 J
Step 3: Calculate the remaining heat energy. Since the ice and steam are at the same temperature after mixing, the remaining heat energy can be calculated by subtracting the heat absorbed by the ice from the heat released by the steam.
Remaining heat energy = Heat released by steam - Heat absorbed by ice Remaining heat energy = 13560 J - 2004 J Remaining heat energy = 11556 J
Step 4: Convert the remaining heat energy to the mass of uncondensed steam. To convert the remaining heat energy to the mass of uncondensed steam, we can use the formula: Q = m * Lv where Q is the remaining heat energy, m is the mass of uncondensed steam, and Lv is the latent heat of vaporization of steam.
Rearranging the formula, we get: m = Q / Lv
m = 11556 J / 2260 J/gm m ≈ 5.11 gm
Therefore, the mass of uncondensed steam is approximately 5.11 gm.
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