A sample of steam with a mass of .550 g and at a temperature of 1 degrees C condenses into an insulated container holding 4.40 g of water at 5.0 degrees C. Assuming that no heat is lost to the surroundings, what will be the final temperature of the mixture?

Are you sure that is 1 degree C for the steam? I would feel better if it were 100 C. If it is 1 degree C, do you have a heat of vaporization for water at 1 degree C?

It is 100 degrees, I just mistyped!!

(heat released by condensation of steam) + (heat released by cooling 100 C water to Tfinal) + (heat absorbed by 4.40 g H2O at 5.0 degrees C moving up to Tf) = 0

[0.550 x heat vap] + (0.550 x specific heat water x (Tfinal-Tinitial)] + [4.40 x specific heat water x (Tfinal-Tinitial)]= 0
Ti for the steam is 100; Ti for the 4.40g H2O is 5.0C

Sorry, I never saw this response so I reposted it correctly! But I do have a question. I used 2257 kJ/kg for the heat vap. and 2.08 J/g*K for the specific heat of water. I'm still getting the problem wrong, am I plugging in the wrong numbers??

To determine the final temperature of the mixture, we can use the principle of energy conservation. The heat lost by the steam during condensation is equal to the heat gained by the water.

First, we need to calculate the heat lost by the steam during condensation. We can use the formula:

Q = m * c * ΔT

where:
Q is the heat lost
m is the mass of the steam
c is the specific heat capacity of steam (assumed constant)
ΔT is the change in temperature

Given:
m = 0.550 g

The specific heat capacity of steam is different from water, but since we assume the steam condenses into water and energy is conserved, we can use the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.

ΔT = 1°C (steam condenses from 1°C to the final temperature)

Q = (0.550 g) * (4.18 J/g°C) * (1°C)
Q = 2.299 J

Next, we need to calculate the heat gained by the water:
Q = m * c * ΔT

where:
Q is the heat gained
m is the mass of the water
c is the specific heat capacity of water (assumed constant)
ΔT is the change in temperature

Given:
m = 4.40 g
c = 4.18 J/g°C (specific heat capacity of water)
ΔT = Tf - 5.0°C (change in temperature, where Tf is the final temperature of the mixture)

We can set up the equation:

2.299 J = (4.40 g) * (4.18 J/g°C) * (Tf - 5.0°C)

Simplifying the equation:

2.299 J = 18.352 g * (Tf - 5.0°C)

Dividing both sides by 18.352:

0.125 J/g°C = Tf - 5.0°C

Adding 5.0°C to both sides:

Tf = 0.125 J/g°C + 5.0°C

Tf = 5.125°C

Therefore, the final temperature of the mixture will be approximately 5.125°C.