If 10g of steam at 110 c is pumped into an insulated vessel containing 100g of water at 20 c, what will be the equilibrium temperature of the mixture? spheat H2O(l)=4.18J/g*c, spheat h2O(g)=2.03J/g*c, heat of vap H2O(l)=2.260kJ/g

Answer 1

loss of heat moving steam from 110 to 100 is
q1 = mass steam x specific heat steam x (Tfinal-Tinitial) . Tf = 100; Ti = 110

q2 = loss of heat on condensing steam at 100 to liquid water at 100.
q2 = mass steam x heat vaporization.

I would do this in two steps.
First step:
q1 + q2 = mass water x specific heat x (Tf-Ti) where mass water is 100 and Ti= 100 C.
Then take that solution, use it as new Ti and use
mass 10g H2O x specific heat x (Tf - 100) + mass 100 g H2O x specific heat x (Tf-new Ti) = 0
Solve for Tf. I think something like 77 C is the answer.