When 165 mL of water at 22 degrees Celsius is mixed with 85 mL of water at 82 degrees Celsius, what is the final temperature? Assume that no heat is lost to the surroundings, and the specific heat of water is at 4.184 J/g C. The density of water is 1.00 g/mL.
Solution:
q=mc∆t
For water q1 = q2
m1c∆t1 = m2c∆t2
If the density of water is 1.00 g/mL then m1 = 165 g and m2 = 85 g
165 × (tf - 22) = 85 × (82 - tf)
tf = 42.4 degrees Celsius
Comments
Leave a comment