with delta(E) = 0 = Qshoe + Qsteam
-Qshoe = Qsteam
208076.96J = L Msteam with L being the latent heat of vaporization = 2256 kJ/kg = 2256000 J/kg
Msteam = .0922kg = .0922L
I think I must be missing something here
m = .5kg
c = 4187 J/(kg*K)
M = .7kg
C = 460.548 J/(kg*K)
and 70C + 273K = 343K
I'm getting an outrageously high change..
So, setting the two Q's equal, I found delta(T) of the shoe to be = -(mc 70C)/(MC)
m - mass of the water
c - specific heat of water
M - mass of the shoe
C - specific heat of iron
Do they both end up at 100C? So the horseshoe would drop 1100C? But then for the heat absorbed by the water, I am not sure what it equals. Is it related to the horseshoe's Q = mc delta(T)? As in, I solve that with delta(T) = 1100C and that's how much heat the water absorbs?
The water needs to rise to 100C first, right? I'm just generally confused how to work the equation up to that point, and then what I need to do to find how much water boils off.