physics help: please help me answer

[StevenYoo]

hi!

5.0 g of steam at 100.0 deg C are passed into 95 g of water at 10.0 deg C. What will be the final temperature of the water?

i thought that maybe the heat lost by the steam will equal the heat gained by the water, so i did:

q = mc delta(T)

m(steam) x c(steam) x (Tf - Ti) = m(water) x c(water) x (Tf - Ti)

(5g) x (0.418 cal/g/C) x (Tf - 100C) = (95g) x (1 cal/g/C) x (Tf - 10C)

if you solve for Tf, you get 7.67 deg C, which canNOT be right...

what am i doing wrong?

it's probably a REALLY simple problem, but i can't get it! oy!

 

[minendo]

You have to use steam tables.

 

[silverpig]

You have a latent heat you have to account for as well. The steam gives up energy to the water and remains steam at 100C for a while. Once the steam condenses, then you have an equation dealing with 5g of water at 100 and 95 g of water at 10.

Also, you should always use kelvin.

 

[Howard]

Does this have anything to do with ironing?

 

[minendo]

Originally posted by: silverpig
You have a latent heat you have to account for as well. The steam gives up energy to the water and remains steam at 100C for a while. Once the steam condenses, then you have an equation dealing with 5g of water at 100 and 95 g of water at 10.

Also, you should always use kelvin.

Steam condenses pretty quick since it is direct steam injection.

 

[lowfatbaconboy]

Originally posted by: Howard
Does this have anything to do with ironing?

wrong again howard...... ironing * pi
is the correct way to do it or something

 

[pray4mojo]

Where did you get your c's from?

 

[silverpig]

Originally posted by: minendo

Originally posted by: silverpig
You have a latent heat you have to account for as well. The steam gives up energy to the water and remains steam at 100C for a while. Once the steam condenses, then you have an equation dealing with 5g of water at 100 and 95 g of water at 10.

Also, you should always use kelvin.

Steam condenses pretty quick since it is direct steam injection.

Didn't say it'd take long... but you still have that latent energy to account for.

Link explaining latent heat

How much heat does it take to get water to change state? If the water is at a temperature of 100 degrees C (that is, the boiling point, or 212 degrees F) it takes an additional 540 calories of heat to convert one gram of water from the liquid state to the vapor state. When the vapor converts to the liquid state, 540 calories of energy will be released per gram of water.

 

[DrPizza]

There really is a simple formula for it, but I always forget the simple formula... (edit: ) I don't believe that memorizing formulas aids in understanding physics...
Here's a method that may make sense for you...
Pick a temperature below what you expect.. say, 10 degrees C
Then calculate how many calories of energy you have that will heat the system above 10 degrees Celsius.
For the water, you have 0 extra calories...
For the steam, you have the calories to heat 5 grams from 10 to 100, plus the calories to evaporate it.

That's your total number of calories of energy that you're adding to a total of 100 grams of water at 10. Figure out the final temperature.

2nd edit... didn't notice the temp of the water was 10 C, I originally thought I had read 95..

 

[silverpig]

540 * 5 = 2700 cal from the latent heat of the steam

There's your starting point. Redo the question as you have done, add this energy in, and change the 5g of steam to 5g of water on the left.