Recent content by madison222s

Oh of course. Ok I got it, thanks!

No, nevermind. Just one. For the ice water. Q4= micecice(22.7-0)

So what I really need is 2 extra equations then?

So then its Q3=(mice+mwater)c(Tf-Ti) =(.3kg + mice)(4190)(22.7-78.6) =(.3kg + mice)(-234221) = -70266.3 - 234221mice And then add this Q3 to the others and set =0?

Then would Q3 have to include both the masses?

Homework Statement An insulated beaker with neglible mass contains .3kg of water at a temp of 78.6 C. How many kg of ice at -19.6 C must be dropped into the water to make Tf= 22.7C? Specific heat for water= 4190 J/kgK specific heat for ice= 2100 J/kgK heat of fusion for water= 334kJ/kg...