calculate the amount of heat ( in kilojoules ) required to convert 74.6 g of water to steam at 100 C??
2007-01-22 09:46:31 · 2 answers · asked by mazoqo 1 in Science & Mathematics ➔ Chemistry
First, translate grams into moles. Water has 18.01488 g/mol, so 74.6 g divided by 18.01488 g/mol is about 4.141 moles of water. Since the water is already at 100 degrees Celsius, just multiply the moles by the heat of vaporization of water, which is 40.65 kJ/mol. 4.141 mol times 40.65 kJ/mol is 168 kJ. Hope this helps.
2007-01-22 10:23:59 · answer #1 · answered by Wesley B 2 · 0⤊ 0⤋
You never stated the starting temperature of the water.
Assuming the temperature is 25°C.
E - Energy.
â²t - Difference in temperature.
shc - Specific heat capacity of water.
E = â²t * shc * mass
E = (100°-25°)(4180)(74â6 x 10^-3)
E = (75)(4180)(0â0746)
E = 23387â1
E â 23â39 KJ
2007-01-22 18:03:19 · answer #2 · answered by Brenmore 5 · 0⤊ 0⤋