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100 g of water is supercooled to - 10^@C...

100 g of water is supercooled to - `10^@C.` At this point, due to same disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperautre of the resultant mixture and how much mass would freeze ?`[s_w = 1 cal//g// .^@ C and L_(Fusion)^(w) = 80 cal//g]`

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To solve the problem, we need to determine the temperature of the resultant mixture and the mass of water that freezes when 100 g of supercooled water at -10°C is disturbed. ### Step-by-Step Solution: 1. **Identify the Initial Conditions**: - Mass of water, \( m = 100 \, \text{g} \) - Initial temperature of supercooled water, \( T_i = -10 \, \text{°C} \) - Final temperature of the mixture after freezing, \( T_f = 0 \, \text{°C} \) (since water freezes at 0°C) ...
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