# Latent Heat

1. Nov 29, 2006

### Meowzers

1. The problem statement, all variables and given/known data

Steam at 100°C is injected into 1.8 kg of water at 22°C in a well-insulated container, where it condenses and mixes with the existing water, reaching thermal equilibrium. If the final temperature of the well-mixed water is 25°C, what is the mass of the injected steam?

2. Relevant equations

Q=(latent heat of vaporization)*m
Specific heat: Q=cmT

3. The attempt at a solution

Latent heat of vaporization of water: 2.26*10^6 (J/kg)

Well, I thought that Q is the amount of heat that is being released as the steam is condensed into water.
Steam: Q=2.26*10^6 * m (m is what I'm trying to find)

The amount of heat being released from the steam condensing to water is the amount of heat that raises the temperature of the water from 22C to 25C.
Water: Q=cmT
Q=(4178 J/kg*K)*(1.8 kg)*(25-22)
Q=22561.2 J

Then, I set the Q's equal to each other to get 0.009, which is incorrect.

Do I need to account for the change of the steam's temperature frorm 100C to 25C? Or add in the unknown mass (I got roughly the same answer 0.010038 kg)?

2. Nov 29, 2006

### OlderDan

You do have to consider the temperature change of the condensed steam from 100C to 25C. I don't think you did your revised calculation correctly.

3. Nov 29, 2006

### Meowzers

So, I use the specific heat equation to calculate the heat released by the steam?

Q=cmT (steam)
Q=(4178 J/kg*K)*(m)*(100-25)
Q=13350*m

Then, I just added the 2 Q's together and set it equal to the specific heat stuff of the water.

(2.26*10^6)(m) + 313350m = (4178)(1.8)(3)
2573350m = 22561.2
m = 0.00877 kg

Is this the correct approach? Or should I have not added the Q's together?

4. Nov 29, 2006

### Meowzers

Nevermind, I got it. Thanks for the response!

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