Chemistry Reference
In-Depth Information
The change in the temperature of the water is The
change in the temperature of the metal is Note
that both the water and the metal wind up at Note also that the change in
temperature of the metal is negative; it cooled down. Substituting into the equa-
tion yields
15.4°C
10.3°C
5.1°C.
15.4°C
67.7°C
52.3°C.
15.4°C.
(27.6 g) (4.184 J/g
#
°C) (5.1°C)
0
(12.5 g) (
c
metal
)(
52.3°C)
0.90 J/g
#
°C
c
metal
J/g
#
°C
The metal in the table with the heat capacity closest to 0.90
is aluminum.
Practice Problem 14.10
A 186-g sample of a metal at is
placed in 251 g of water at and the final temperature of the system is
Calculate the specific heat of the metal. Which of the metals in Table 14.4
could it be?
74.0°C
18.0°C,
26.4°C.
EXAMPLE 14.11
0.892 J/g
#
°C)
Calculate the final temperature after 6.79 g of a metal
(
c
at
67.4°C
is placed in 196 g of water at
19.6°C.
Solution
Heat
0
(
m
water
)(
c
water
)(
¢
t
water
)
(
m
metal
)(
c
metal
)(
¢
t
metal
)
Let
t
f
represent the final temperature of
both
the metal and the water:
0
(
m
water
)(
c
water
)(
t
f
t
water
)
(
m
metal
)(
c
metal
)(
t
f
t
metal
)
Substituting the known quantities gives
(196 g)(4.184 J/g
#
°C)(
t
f
(6.79 g)(0.892 J/g
#
°C) (
t
f
0
19.6°C)
67.4°C)
0
(196) (4.184) (
t
f
19.6°C)
(6.79) (0.892) (
t
f
67.4°C)
The equation is simplified using the distributive property of algebra, each prod-
uct having three significant digits:
The algebraic simplification is
based on the equality
0
820.
1t
f
16,0
7
0°C
6.05
7t
f
408.
2
°C
wx(y
z)
wxy
wxz
826.
2t
f
16,4
8
0°C
t
f
19.9°C
Practice Problem 14.11
Calculate the final temperature after 11.3 g of
a metal
0.44 J/g
#
°C)
(
c
at
60.0°C
is placed in 97.2 g of water at 19.0°C.
Change of Phase Calculations
When energy is added to or removed from a pure substance and that substance
changes phase
as a result, the
temperature does not change.
That is, when a
pure substance melts, solidifies, sublimes, evaporates, or condenses, it does so
