Exactly 11.1 mL of water at 24.0 degree C are added to a hot iron skillet. All o
ID: 494728 • Letter: E
Question
Exactly 11.1 mL of water at 24.0 degree C are added to a hot iron skillet. All of the water is converted in steam at of the pan is 1.35 kg and the molar heat capacity of iron is 25.19 J(mol middot degree C). What is the temperature change of the skillet? You actually don't need to know the initial temperature of the iron skillet since you are only asked for delta T, not the actual final T. All you need to do is calculate the total amount of heat needed to BOTH raise the temperature of the water to 100 degree C and then boil it. This total amount of heat came FROM the skillet and must cause the temperature of the skillet to decrease.Explanation / Answer
Heat capacity of water = 4.184J/g-Co
Latent heat of vapourisation of water = 2260J/g
Density of water = 1g/mL
Volume of water given = 11.1mL
Density = Mass/Volume
1 = Mass/11.1
Mass = 11.1g
Temperature of water in skillet is changed from 24oC to 100oC and then its liquid state is changed to steam.
Change in temperature for water = 100 - 24 =76oC
Energy required to change temperature of water = Heat capacity of water *change in temperature*mass of water
=4.184*76*11.1 =3529.6224J
Energy required to change liquid state of water to steam = Latent heat of vapourisation *Mass of water
=2260*11.1 = 25086J
Total energy consumed = 3529.6224 + 25086 = 28615.6224J
Energy required in the process will be acquired by from iron skillet.
Mass of iron pan = 1.35kg
Moles of iron in 1.35kg = (1.35*1000)/56 = 24.286
Molar heat capacity of iron pan = 25.19J /mol-oC
Energy released by skillet = Change in temperature for pan *molar heat capacity*moles of iron
=Change in temperature *25.19*24.286
Energy consumed by water = Energy released by iron skillet
28615.6224 = Change in temperature for pan *25.19*24.286
Change in temperature for iron skillet or pan = 46.775oC
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