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physics homework help....?
If 2.80x10^5 J of energy is supplied to a flask of liquid oxygen at -183 C, how much oxygen can evaporate?
3 Answers
- Charith PLv 41 decade agoFavorite Answer
-183 C is also the boiling temperature of oxygen, this energy will go into changing the state of oxygen from a liquid to a gas. This is known as the heat of vaporisation.
According to wiki, this value is 6.82 kJ·mol−1 for oxygen.
We can work out how many moles of oxygen can evaporate by working how how many times of this energy we provide:
2.8x10^5/6.82x10^3 = 41.06 moles.
Each mole of oxygen is 32 grams so multiplying this value by 32, we get the mass of the vaporised oxygen - 41.06 x 32 = 1313.78 grams
- 1 decade ago
Q=mL
2.80 x 10^5 J = m(L)
L= heat of vaporization for oxygen= 210 kJ/kg
2.80 x 10^5 J = m(210 x 10^3 J/kg)
= 1.3kg
:)
