If an ice cube weighing 25.0 g with an initial
temperature of -7.5 °C is placed in an insulated cup
containing 200 ml of water that has an initial
temperature of 15.7 °C, what is the final temperature of
the water?
(Specific heat capacity of water is 4.18 J/°C.g.
Density of water is 1 g/ml)
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Ответ:
11
∘
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
∘
C
to liquid at
0
∘
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
=
−
q
3
(
1
)
, where
q
1
- the heat absorbed by the solid at
0
∘
C
q
2
- the heat absorbed by the liquid at
0
∘
C
q
3
- the heat lost by the warmer water sample
The two equations that you will use are
q
=
m
⋅
c
⋅
Δ
T
, where
q
- heat absorbed/lost
m
- the mass of the sample
c
- the specific heat of water, equal to
4.18
J
g
∘
C
Δ
T
- the change in temperature, defined as final temperature minus initial temperature
and
q
=
n
⋅
Δ
H
fus
, where
q
- heat absorbed
n
- the number of moles of water
Δ
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
⋅
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
∘
C
to liquid at
0
∘
C
?
q
1
=
5.551
moles
⋅
6.01
kJ
mole
=
33.36 kJ
This means that equation
(
1
)
becomes
33.36 kJ
+
q
2
=
−
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
⋅
c
⋅
Δ
T
sample
=
−
m
water
⋅
c
⋅
Δ
T
water
More specifically, you have
33.36 kJ
+
100.0
g
⋅
4.18
J
g
∘
C
⋅
(
T
f
−
0
)
∘
C
=
−
650
g
⋅
4.18
J
g
∘
C
⋅
(
T
f
−
25
)
∘
C
33.36 kJ
+
418 J
⋅
(
T
f
−
0
)
=
−
2717 J
⋅
(
T
f
−
25
)
Convert the joules to kilojoules to get
33.36
kJ
+
0.418
kJ
⋅
T
f
=
−
2.717
kJ
⋅
(
T
f
−
25
)
This is equivalent to
0.418
⋅
T
f
+
2.717
⋅
T
f
=
67.925
−
33.36
T
f
=
34.565
0.418
+
2.717
=
11.026
∘
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
∘
C
Explanation:
Ответ:
2 moles KClO₃ > 3 moles O₂
3.98 moles KClO₃ > ?
moles O₂ = 3.98 x 3 / 2
moles O₂ = 11.94 / 2
= 5.97 moles of O₂
hope this helps!