I object to the final equilibration step.
30g of H2O@ 50 C are mixed with 1g of steam @110 C. After final equilibration, the temp is…? (Heat condensation=540cal/g, specific heat of steam=0.5cal/g, sp.heat H2O = 1cal/g)
The answer is about 70 C
The last calculation is: deltaQ of H2O = deltaQ of steam
(30)(1g)(Tfinal-T) = (1)(0.5)(100-Tfinal)
I’m truckin’ along, on top of the world like Leonardo until the part about the specific heat = 0.5 comes in. See, I think that by the time the steam reaches 100C I’ve already converted it into water… so why would I use the specific heat of steam at the end again?
