Change Of State Problems Chemistry I Honours Page 2
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1
2o
o
5. When 15.00 grams of cold water at 12.0
C are added to 50.0 grams of water at 85.0
C, what will be the
final temperature of the mixture?
mC t = mC t
(heat lost – heat gained)
since the water is in the same state, the specific heat (C) can be removed.
o
o
(15.00 g)(X – 12.0
) = (50.0 g)(85.0
C – X)
15.00 X – 180 = 4250 – 50.0 X
o
X = 68.1
C
65.0 X = 4430
6. What is the specific heat of a metal if 75.00 grams of the metal causes 125.00 grams of water to increase
o
o
o
in temperature from 20.00
C to 40.00
C? (The original temperature of the metal was 88.8
C; its final
o
temperature was 40.0
C.)
mC t = mC t
(heat lost by metal – heat gained by water)
o
o
(125.0 g)(4.184 J/g-
C)(20.00
)
o
2.86 J/g-
C
then C
= ------------------------------------------- =
metal
o
(75.00 g)(48.8
)
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