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?
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...