Answer:
11
β
C
Explanation:
As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at
0
β
C
to liquid at
0
β
C
.
The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have
q
1
+
q
2
=
β
q
3
(
1
)
, where
q
1
- the heat absorbed by the solid at
0
β
C
q
2
- the heat absorbed by the liquid at
0
β
C
q
3
- the heat lost by the warmer water sample
The two equations that you will use are
q
=
m
β
c
β
Ξ
T
, where
q
- heat absorbed/lost
m
- the mass of the sample
c
- the specific heat of water, equal to
4.18
J
g
β
C
Ξ
T
- the change in temperature, defined as final temperature minus initial temperature
and
q
=
n
β
Ξ
H
fus
, where
q
- heat absorbed
n
- the number of moles of water
Ξ
H
fus
- the molar heat of fusion of water, equal to
6.01 kJ/mol
Use water's molar mass to find how many moles of water you have in the
100.0-g
sample
100.0
g
β
1 mole H
2
O
18.015
g
=
5.551 moles H
2
O
So, how much heat is needed to allow the sample to go from solid at
0
β
C
to liquid at
0
β
C
?
q
1
=
5.551
moles
β
6.01
kJ
mole
=
33.36 kJ
This means that equation
(
1
)
becomes
33.36 kJ
+
q
2
=
β
q
3
The minus sign for
q
3
is used because heat lost carries a negative sign.
So, if
T
f
is the final temperature of the water, you can say that
33.36 kJ
+
m
sample
β
c
β
Ξ
T
sample
=
β
m
water
β
c
β
Ξ
T
water
More specifically, you have
33.36 kJ
+
100.0
g
β
4.18
J
g
β
C
β
(
T
f
β
0
)
β
C
=
β
650
g
β
4.18
J
g
β
C
β
(
T
f
β
25
)
β
C
33.36 kJ
+
418 J
β
(
T
f
β
0
)
=
β
2717 J
β
(
T
f
β
25
)
Convert the joules to kilojoules to get
33.36
kJ
+
0.418
kJ
β
T
f
=
β
2.717
kJ
β
(
T
f
β
25
)
This is equivalent to
0.418
β
T
f
+
2.717
β
T
f
=
67.925
β
33.36
T
f
=
34.565
0.418
+
2.717
=
11.026
β
C
Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be
T
f
=
11
β
C
Explanation:
