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How should 1 kg of water at 50^(@)C be d...

How should `1 kg` of water at `50^(@)C` be divided in two parts such that if one part is turned into ice at `0^(@)C`. It would release sufficient amount of heat to vapourize the other part. Given that latent heat of fusion of ice is `3.36xx10^(5) J//Kg`. Latent heat of vapurization of water is `22.5xx10^(5) J//kg` and specific heat of water is `4200 J//kg K`.

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To solve the problem, we need to divide 1 kg of water at 50°C into two parts: one part (mass \( x \) kg) will be turned into ice at 0°C, and the other part (mass \( 1 - x \) kg) will be vaporized. We will use the principles of calorimetry to find the value of \( x \). ### Step 1: Calculate the heat released by the part that turns into ice 1. **Cooling the water from 50°C to 0°C**: The heat released when the mass \( x \) kg of water is cooled from 50°C to 0°C can be calculated using the formula: \[ Q_1 = m \cdot s \cdot \Delta T ...
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