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lina2011 [118]
3 years ago
6

Water's heat of fusion is 80. cal/g , its specific heat is 1.0calg⋅∘C, and its heat of vaporization is 540 cal/g .

Chemistry
1 answer:
andrey2020 [161]3 years ago
6 0

Answer:

Q1=(330g)*(80cal/g)=26,400cal

Q2=(430cal)*(1cal/(g-c)*(100c=43,000cal

Q3=(430g)*(540 cal/g)= 232,200cal

Total heated absorbed is

<h3> Q1+Q2+Q3= 301,600 or 301.6 kcal</h3>

Explanation:

# I hope it's help

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