Recent content by Shmiam

Awesome! Thank you very much for all of the guidance, I think it helped me to understand the concepts here much more

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

At equilibrium it would match the temperature of the water, so Q = MC(373K-1018.433K), which comes out to be -208076.96J?

The water goes through its phase change from liquid to steam, and it stops if it all evaporates or if the system reaches thermal equilibrium

In that case, the horseshoe loses 454.567K

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

Oh, then the horseshoe must have lost Q

I'd have to say I don't know

What I meant by that was whether the horseshoe lost mc (70C) or if that's the wrong way to think about it

Would its temperature change also equal 70C?

c is the specific heat of water and m is the mass, which, using density, would be .5kg

Would that be = mc(70C)?

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.