Asked by hamda
How much energy in form of heat is required to convert a 10.0-g sample of ice initially at -5.00C to steam at 120C ? Specific heats (in J/g • ºC) of ice, water liquid and steam are 2.108, 4.184 and 1.996 respectively
Answers
Answered by
Anonymous
ice from -5 to 0
10 * 2.108 * 5 =
melt ice
10 * heat of fusion of water =
heat water from 0 to 100
10 * 4.184 * (100-0) =
boil water
10 * heat of vaporization of water =
superheat steam
10 * 1.996 * (120 - 100) =
add them
10 * 2.108 * 5 =
melt ice
10 * heat of fusion of water =
heat water from 0 to 100
10 * 4.184 * (100-0) =
boil water
10 * heat of vaporization of water =
superheat steam
10 * 1.996 * (120 - 100) =
add them
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