Mix steam with colder water, what is final temp?

AI Thread Summary
The problem involves mixing 100 g of steam at 100°C with 1 kg of water at 10°C to find the final temperature of the mixture. The phase change of condensation from steam to water must be considered, resulting in energy loss calculated using the latent heat of vaporization. After the phase change, heat transfer occurs between the warmer water and the colder water until thermal equilibrium is reached. The calculations indicate that the final temperature should be around 67°C, despite an initial miscalculation suggesting 60°C. Correctly applying the heat transfer equations is crucial for accurate results.
SunshineCat
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Homework Statement


100 g of steam at 100C is mixed with 1 kg of water at 10C. What is the final temperature of the mixture? (in C)
heat capacity of water: cw = 4.186 kJ kg-1 K−1
latent heat of vapourisation for water: Lv = 2256 kJ kg−1
latent heat of fusion for water: Lf = 334 kJ kg−1

Homework Equations


Q=mL
Q=mcΔT

The Attempt at a Solution


I think that there needs to be a phase change of condensation, 0.1kg of 100C steam to 0.1kg of 100C water, so this will be Q=mL and will be an energy loss
The energy lost will go to the colder water
After this, energy will keep flowing from the 100C water to 10C water until they reach the same temperature, energy gain and loss will be Q=mcΔT
I can't seem to get the calculation right, I keep getting 60C when the answer is 67C

ΔQ (steam) = ΔQ(water)
Q=mcΔT = mL + mcΔT
0.1x4.186x(T-100) = 1x4.186x(T-10) - 0.1x225.6
225.6=3.77T
T=60C?
 
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SunshineCat said:
0.1x4.186x(T-100) = 1x4.186x(T-10) - 0.1x225.6

That should be 0.1 * 2256
 
CWatters said:
That should be 0.1 * 2256
Oh yeah, that was a typo I think
 
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