A 0.598 g sample of steam at 104.8 ∘C is condensed into a container with 4.91 g of water at 16.7 ∘C. What is the final temperature of the water mixture if no heat is lost?

Answer 1

The heat lost by the steam MUST equal the heat gained by the water. It's best to do this in steps:

Step 1: heat lost when steam goes from 104.8 to 100 degrees

q = mC∆T = (0.598 g)(1.89 J/gº)(4.8º) = 5.43 J

Step 2: heat lost when steam at 100º changes to liquid @ 100º

q = m∆vap = (0.598 g)(2260 J/g) = 1351 J

Step 3: heat lost when liquid @ 100 cools to a final temperature, Tf:

q = mC∆T = (0.598 g)(4.184 J/gº)(100º - Tf) = 250J - 25Tf

Step 4: Total heat lost by the steam:

5.43 J + 1351 J + 250 J - 25Tf

Step 5: heat gained by water:

q = mC∆T = (4.91 g)(4.184 J/gº)(Tf - 16.7º) = 20.7Tf - 345

Step 6: Set heat lost by steam equal to heat gained by water:

5.43 J + 1351 J + 250 J - 25Tf = 20.7Tf - 345

Step 7: solve for Tf (final temperature):

45.7Tf = 1951

Tf = 42.7º

(be sure to check all of the math. also, if you use different values for ∆Hf and C, answer may be different)