- #1
GodsChild086
- 22
- 0
I'm really stuck on this question:
What will be the final temperature if 50.0 g of steam at 100 degrees celsius is mixed with 600 mL of water at 10.0 degrees celsius? (Answer: 58.3 degrees celsius)
So far I've got:
heat lost (steam) = heat gained (water)
nH + mc(delta)t = mc(delta)t
Sorry I don't know how to make the delta symbol.
(500g/18.02g/mol)(40.65 kJ/mol) + 0.500 kg(2.02 kJ/kg*C)(100-t) = 0.600 kg(4.19 kJ/kg*C)(t-10)
When I calculated everything, I didnt' get 58.3*C, so maybe someone can tell me where I went wrong in either my formulas or calculations. Thanks.
What will be the final temperature if 50.0 g of steam at 100 degrees celsius is mixed with 600 mL of water at 10.0 degrees celsius? (Answer: 58.3 degrees celsius)
So far I've got:
heat lost (steam) = heat gained (water)
nH + mc(delta)t = mc(delta)t
Sorry I don't know how to make the delta symbol.
(500g/18.02g/mol)(40.65 kJ/mol) + 0.500 kg(2.02 kJ/kg*C)(100-t) = 0.600 kg(4.19 kJ/kg*C)(t-10)
When I calculated everything, I didnt' get 58.3*C, so maybe someone can tell me where I went wrong in either my formulas or calculations. Thanks.