Asked by Anonymous
A sample of steam with a mass of 0.524g and at a temperature of 100 ∘C condenses into an insulated container holding 4.20g of water at 6.0∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C) Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?
Answers
Answered by
DrBob222
heat lost by condensing steam + heat lost by condensed steam @ 100 cooling to Tf + heat gained by cool(6C) water = 0
-(2259 J/g x 0.524) + [(0.524 x 4.18 x (Tfinal-Tinitial]) + [(mass cool water x 4.18 x (Tfinal - Tinitial)] = 0
Solve for Tf.
My answer is about 75 C or so.
-(2259 J/g x 0.524) + [(0.524 x 4.18 x (Tfinal-Tinitial]) + [(mass cool water x 4.18 x (Tfinal - Tinitial)] = 0
Solve for Tf.
My answer is about 75 C or so.
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