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Finding the heat of a reaction?
Make the following assumptions:
1) The density of liquid water at 100C is 1.00
g/mL;
2) the vapor is adequately described by the
ideal gas equation;
3) the external pressure is constant at 1.00
atm; and
4) the heat of vaporization of water is 2.26
kJ/g.
What is q for the vaporization of 1.00 g of
liquid water at 1.00 atm and 100C, to the
formation of 1.00 g of steam at 1.00 atm and
100C?
I tried using q=mcT but since there is no change in temperature, I always get 0. It's in the list of possible answers, but I'm seriously doubting myself. Phuk it.
1 Answer
- Caroline MillerLv 78 years agoFavorite Answer
look at the heat of vaporization value. there is no temperature unit. therefore, you don't use temperature in the equation. this the heat required to vaporize 1 g water at the vaporization temperature. since this is 100ºC and all water must be converted to steam at 100ºC, there is no ∆T. if we were to heat the steam to greater than 100ºC. THEN you would use the ∆T
∆H = mass x heat of vaporization
∆H = 1g x 2.26kJ/g = 2.26kJ
