A 0.200 kg ball of glass is heated to a temperature of 82.0 Degree C and then pl
ID: 1539172 • Letter: A
Question
Explanation / Answer
1.) heat gained by ice = mCT = 0.2*0.840*82
energy required to convert ice into water at 0 degree = m*deltaH
m*delta H = 0.2*0.84*82
m*334 = 0.2*0.84*82
m= 0.103 Kg = 103.11 gram
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