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An insulated beaker with negligible mass contains liquid water with a mass of 0.205 kg and a temperature of 67.9 C
How much ice at a temperature of -14.4 C must be dropped into the water so that the final temperature of the system will be 28.0 C?
Take the specific heat of liquid water to be 4190 J/kGK , the specific heat of ice to be 2100 J/kGK , and the heat of fusion for water to be 334 kJ/kG.
mass = kG

2007-04-12 17:53:37 · 1 answers · asked by Anonymous in Science & Mathematics Physics

1 answers

(14.4*2100 + 334,000 + 28*4190)x = 0.205*39.9*4190
x = 34272.105/481560
x = 0.0712 kg

2007-04-12 18:17:08 · answer #1 · answered by Helmut 7 · 0 0

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