Heat Absorbed Changing 2kg Ice to Steam: Calculation

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To calculate the heat absorbed in changing 2 kg of ice at -5°C to steam at 110°C, one must consider the specific heats of ice, water, and steam, along with the heat fusion and vaporization values. The process involves multiple steps: heating the ice to 0°C, melting it to water, heating the water to 100°C, vaporizing it to steam, and finally heating the steam to 110°C. Each step requires applying the respective specific heat or latent heat values. The total heat absorbed is the sum of the heat calculated for each phase change and temperature increase. This calculation demonstrates the comprehensive energy transformation involved in the phase change from ice to steam.
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How much heat is absorbed in changing 2-kg of ice at -5°C to steam at 110.0°C? The Specific heats of ice, liquid water, and steam are, respectively, 2060 J/Kg x K, 4180 J/Kg x K, and 2020 J/Kg x K. The heat fusion of ice is 3.34 x 10^5 J/Kg. The heat of vaporization of water is 2.26 x 10^6 J/kg
 
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