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Find what heat in calories is required to change 52 gg of 0∘C∘C ice to steam at 100∘C∘C. The heat of fusion LfLf for water is 80 cal/gcal/g. The specific heat capacity of water is 1 cal/g⋅∘Ccal/g⋅∘C. The heat of vaporization for water LvLv is 540 cal/gcal/g.

Respuesta :

Answer:

Explanation:

Heat required to melt ice = mass x heat of fusion

= 52 x 80 = 4160 cal

heat required to increase temperature to 100⁰C

mass x specific heat x rise in temperature

= 52 x 1 x 100 = 5200 cals

heat required to boil off all the water = mass x heat of vaporisation

= 52 x 540 = 28080 cal

Total heat required = 4160 + 5200 + 28080

= 37440 cals .

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