%3Cp%3E1.2%20kg%26nbsp%3Bice%26nbsp%3Bat%26nbsp%3B-20%26nbsp%3Bdegrees%26nbsp%3B
ID: 2255694 • Letter: #
Question
%3Cp%3E1.2%20kg%26nbsp%3Bice%26nbsp%3Bat%26nbsp%3B-20%26nbsp%3Bdegrees%26nbsp%3BC%26nbsp%3Bis%26nbsp%3Bheated%26nbsp%3Bto%26nbsp%3B0%26nbsp%3Bdegrees%26nbsp%3BC%26nbsp%3Band%26nbsp%3Bmelted.The%26nbsp%3Bresulting%26nbsp%3Bwater%26nbsp%3Bis%26nbsp%3Bheated%26nbsp%3Bto100%26nbsp%3Bdegrees%26nbsp%3BC%2C%26nbsp%3Bvaporizes%26nbsp%3Band%26nbsp%3BSteam%26nbsp%3Bis%26nbsp%3BfurthER%26nbsp%3Bheated%26nbsp%3Bto300%26nbsp%3Bdegrees%26nbsp%3BC.%26nbsp%3BAssume%26nbsp%3Bthat%26nbsp%3Bthe%26nbsp%3B%26nbsp%3Bprocesses%26nbsp%3Bare%26nbsp%3Bisobaric.%26nbsp%3BCalculate%26nbsp%3Bthe%26nbsp%3Bchange%26nbsp%3Bin%26nbsp%3Benthalpy%26nbsp%3Bin%26nbsp%3Bkilocalories%26nbsp%3Band%26nbsp%3Bjoules.%26nbsp%3B(The%26nbsp%3Bspecific%26nbsp%3Bheats%26nbsp%3Bof%26nbsp%3Bice%2C%26nbsp%3Bwater%2C%26nbsp%3Band%26nbsp%3Bsteam%26nbsp%3Bare%26nbsp%3B0.55%2C1.00%2C%26nbsp%3Band%26nbsp%3B0.48%26nbsp%3Bkcal%2Fkg%26nbsp%3B0%26nbsp%3Bdegrees%26nbsp%3BC%2C%26nbsp%3Brespectively.%3C%2Fp%3EExplanation / Answer
specific heat of water is 4.186 kJ/kgC
heat of fusion of ice is 334 kJ/kg
E = [334 kJ/kg x 0.2 kg] + [ 4.186 kJ/kgC x 0.2 kg x 60C ]
E = 66.8 + 50.23 kJ = 117.03 kJ
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