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1Q11.04 Homework. Unanswered The temperature of a 0.555 kg block of ice is lowered to -120°C. Heat (0.5159 MJ) is then transf

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Heat transferred = 0:5159 MJ = 0.5159 810 I 2 ms DT sa = specific heat of water 4.18 J/gme ДТ - mas = 5.159 X 10 3 555 gm x 4Дт - 2 2 2 - зв. т, - т. - 22.2 - звc 2 - т, + 2 2 2 . Bв с - - | 20 c +222-38 с Пт, то 2 - 38 с

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