Chem need help?
The question is 50g of steam at 165 degree is added to 200g of ice at -45 degree, what will be the final temperature of the system?
- 1 decade agoFavorite Answer
you have to use the formula q=mc(final temp - initial temp)
The steam will lose heat (q) and therefore will be negative.
-q = mc(final temp - initial temp)
The ice will gain heat and will still be q = mc (final - initial).
The heat loss for the steam is equal to the heat gained from the ice so you can set the two formulas equal together, so...
- mc (final - initial) = mc (final - initial)
^^^^^ steam ^^^^^^^ ice
So now we can put in the numbers.
The specific heat (c) of steam is 1.87 J/g Celsius
The specific heat of ice is 2.06 J/g Celsius
The final temperature of the two will be the same.
-50 g * 1.87 J/gC (final-165C) = 200 g * 2.06 J/gC (final - (-45C))
Plug some numbers into your calculator and you get
-93.5 J/C (final-165C) = 412 J/C (final + 45C)
Divide both sides by 95.5 J/C and you get
final - 165C = -4.406 (final + 45 C)
Then distribute the -4.406 and you get
final -165 C = -4.046*final - 198.29 C
Combine the like terms and you get
5.406*final = -33.29 C
Divide both sides by 5.406 and you get your answer of
final temperature = -6.158 C
Gotta Love Thermo
- OracleLv 41 decade ago
After what period of time?