119k views
4 votes
Find the final temperature when 10 grams of aluminum at 130 c mixes with 200 grams of water at 25 c. assume no water is lost as water vapor. specific heat values (c) for aluminum= 0.901 j/g c and water 4.18 j/g c

User B Kansara
by
4.3k points

1 Answer

8 votes

Answer:

About 26.12 C°

Step-by-step explanation:

Following q = mc∆t, assuming q is the same
(10g)(130 - T)(0.901j/g*c) = (200.0g)(T - 25)(4.18j/g*c)
T = 26.12

User Akkonrad
by
4.1k points