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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

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Answer:

About 26.12 C°

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

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