527
18-4 ABSOR PTION OF H EAT
Q 3 ϭ c liq m(T f Ϫ T i )
ϭ (4186.8 J/kg иK)(0.720 kg)(15ЊC Ϫ 0°C)
ϭ 45 217 J Ϸ 45.22 kJ.
Total: The total required heat Q tot is the sum of the
amounts required in the three steps:
Q tot ϭ Q 1 ϩ Q 2 ϩ Q 3
ϭ 15.98 kJ ϩ 239.8 kJ ϩ 45.22 kJ
Ϸ 300 kJ.
(Answer)
Note that most of the energy goes into melting the ice rather
than raising the temperature.
(b) If we supply the ice with a total energy of only 210 kJ (as
heat), what are the final state and temperature of the water?
KEY IDEA
From step 1, we know that 15.98 kJ is needed to raise the
temperature of the ice to the melting point. The remaining
heat Q rem is then 210 kJ Ϫ 15.98 kJ, or about 194 kJ. From
step 2, we can see that this amount of heat is insufficient to
melt all the ice. Because the melting of the ice is incomplete,
we must end up with a mixture of ice and liquid; the temperature of the mixture must be the freezing point, 0ЊC.
Calculations: We can find the mass m of ice that is melted by
the available energy Q rem by using Eq. 18-16 with L F :
Thus, the mass of the ice that remains is 720 g Ϫ 580 g, or
140 g, and we have
580 g water and 140 g ice, at 0ЊC.
(Answer)
m ϭ
Q rem
L F
ϭ
194 kJ
333 kJ/kg
ϭ 0.583 kg Ϸ 580 g.
Sample Problem 18.04 Heat to change temperature and state
(a) How much heat must be absorbed by ice of mass m ϭ
720 g at Ϫ10ЊC to take it to the liquid state at 15ЊC?
KEY IDEAS
The heating process is accomplished in three steps: (1) The
ice cannot melt at a temperature below the freezing
point — so initially, any energy transferred to the ice as
heat can only increase the temperature of the ice, until 0ЊC is
reached. (2) The temperature then cannot increase until all
the ice melts — so any energy transferred to the ice as heat
now can only change ice to liquid water, until all the ice melts.
(3) Now the energy transferred to the liquid water as heat can
only increase the temperature of the liquid water.
Warming the ice: The heat Q 1 needed to take the ice from
the initial T i ϭ Ϫ10ЊC to the final T f ϭ 0ЊC (so that the ice
can then melt) is given by Eq. 18-14 (Q ϭ cm ⌬T ). Using the
specific heat of ice c ice in Table 18-3 gives us
Q 1 ϭ c ice m(T f Ϫ T i )
ϭ (2220 J/kg иK)(0.720 kg)[0ЊC Ϫ (Ϫ10ЊC)]
ϭ 15 984 J Ϸ 15.98 kJ.
Melting the ice: The heat Q 2 needed to melt all the ice is
given by Eq. 18-16 (Q ϭ Lm). Here L is the heat of fusion
L F , with the value given in Eq. 18-18 and Table 18-4. We find
Q 2 ϭ L F m ϭ (333 kJ/kg)(0.720 kg) Ϸ 239.8 kJ.
Warming the liquid: The heat Q 3 needed to increase the
temperature of the water from the initial value T i ϭ 0ЊC to
the final value T f ϭ 15ЊC is given by Eq. 18-14 (with the specific heat of liquid water c liq ):
Substituting Eqs. 18-19 through 18-21 into Eq. 18-22 yields
c w m w (T f Ϫ T i ) ϩ C b (T f Ϫ T i ) ϩ c c m c (T f Ϫ T) ϭ 0.
(18-23)
Temperatures are contained in Eq. 18-23 only as differences.
Thus, because the differences on the Celsius and Kelvin
scales are identical, we can use either of those scales in this
equation. Solving it for T f , we obtain
Using Celsius temperatures and taking values for c c and c w
from Table 18-3, we find the numerator to be
(0.0923 cal/g иK)(75 g)(312ЊC) ϩ (45 cal/K)(12ЊC)
ϩ (1.00 cal/g иK)(220 g)(12ЊC) ϭ 5339.8 cal,
T f ϭ
c c m c T ϩ C b T i ϩ c w m w T i
c w m w ϩ C b ϩ c c m c
.
and the denominator to be
(1.00 cal/g иK)(220 g) ϩ 45 cal/K
ϩ (0.0923 cal/g иK)(75 g) ϭ 271.9 cal/CЊ.
We then have
(Answer)
From the given data you can show that
Q w Ϸ 1670 cal,
Q b Ϸ 342 cal,
Q c Ϸ Ϫ2020 cal.
Apart from rounding errors, the algebraic sum of these
three heat transfers is indeed zero, as required by the conservation of energy (Eq. 18-22).
T f ϭ
5339.8 cal
271.9 cal/CЊ
ϭ 19.6ЊC Ϸ 20ЊC.
Additional examples, video, and practice available at WileyPLUS
18-4 ABSOR PTION OF H EAT
Q 3 ϭ c liq m(T f Ϫ T i )
ϭ (4186.8 J/kg иK)(0.720 kg)(15ЊC Ϫ 0°C)
ϭ 45 217 J Ϸ 45.22 kJ.
Total: The total required heat Q tot is the sum of the
amounts required in the three steps:
Q tot ϭ Q 1 ϩ Q 2 ϩ Q 3
ϭ 15.98 kJ ϩ 239.8 kJ ϩ 45.22 kJ
Ϸ 300 kJ.
(Answer)
Note that most of the energy goes into melting the ice rather
than raising the temperature.
(b) If we supply the ice with a total energy of only 210 kJ (as
heat), what are the final state and temperature of the water?
KEY IDEA
From step 1, we know that 15.98 kJ is needed to raise the
temperature of the ice to the melting point. The remaining
heat Q rem is then 210 kJ Ϫ 15.98 kJ, or about 194 kJ. From
step 2, we can see that this amount of heat is insufficient to
melt all the ice. Because the melting of the ice is incomplete,
we must end up with a mixture of ice and liquid; the temperature of the mixture must be the freezing point, 0ЊC.
Calculations: We can find the mass m of ice that is melted by
the available energy Q rem by using Eq. 18-16 with L F :
Thus, the mass of the ice that remains is 720 g Ϫ 580 g, or
140 g, and we have
580 g water and 140 g ice, at 0ЊC.
(Answer)
m ϭ
Q rem
L F
ϭ
194 kJ
333 kJ/kg
ϭ 0.583 kg Ϸ 580 g.
Sample Problem 18.04 Heat to change temperature and state
(a) How much heat must be absorbed by ice of mass m ϭ
720 g at Ϫ10ЊC to take it to the liquid state at 15ЊC?
KEY IDEAS
The heating process is accomplished in three steps: (1) The
ice cannot melt at a temperature below the freezing
point — so initially, any energy transferred to the ice as
heat can only increase the temperature of the ice, until 0ЊC is
reached. (2) The temperature then cannot increase until all
the ice melts — so any energy transferred to the ice as heat
now can only change ice to liquid water, until all the ice melts.
(3) Now the energy transferred to the liquid water as heat can
only increase the temperature of the liquid water.
Warming the ice: The heat Q 1 needed to take the ice from
the initial T i ϭ Ϫ10ЊC to the final T f ϭ 0ЊC (so that the ice
can then melt) is given by Eq. 18-14 (Q ϭ cm ⌬T ). Using the
specific heat of ice c ice in Table 18-3 gives us
Q 1 ϭ c ice m(T f Ϫ T i )
ϭ (2220 J/kg иK)(0.720 kg)[0ЊC Ϫ (Ϫ10ЊC)]
ϭ 15 984 J Ϸ 15.98 kJ.
Melting the ice: The heat Q 2 needed to melt all the ice is
given by Eq. 18-16 (Q ϭ Lm). Here L is the heat of fusion
L F , with the value given in Eq. 18-18 and Table 18-4. We find
Q 2 ϭ L F m ϭ (333 kJ/kg)(0.720 kg) Ϸ 239.8 kJ.
Warming the liquid: The heat Q 3 needed to increase the
temperature of the water from the initial value T i ϭ 0ЊC to
the final value T f ϭ 15ЊC is given by Eq. 18-14 (with the specific heat of liquid water c liq ):
Substituting Eqs. 18-19 through 18-21 into Eq. 18-22 yields
c w m w (T f Ϫ T i ) ϩ C b (T f Ϫ T i ) ϩ c c m c (T f Ϫ T) ϭ 0.
(18-23)
Temperatures are contained in Eq. 18-23 only as differences.
Thus, because the differences on the Celsius and Kelvin
scales are identical, we can use either of those scales in this
equation. Solving it for T f , we obtain
Using Celsius temperatures and taking values for c c and c w
from Table 18-3, we find the numerator to be
(0.0923 cal/g иK)(75 g)(312ЊC) ϩ (45 cal/K)(12ЊC)
ϩ (1.00 cal/g иK)(220 g)(12ЊC) ϭ 5339.8 cal,
T f ϭ
c c m c T ϩ C b T i ϩ c w m w T i
c w m w ϩ C b ϩ c c m c
.
and the denominator to be
(1.00 cal/g иK)(220 g) ϩ 45 cal/K
ϩ (0.0923 cal/g иK)(75 g) ϭ 271.9 cal/CЊ.
We then have
(Answer)
From the given data you can show that
Q w Ϸ 1670 cal,
Q b Ϸ 342 cal,
Q c Ϸ Ϫ2020 cal.
Apart from rounding errors, the algebraic sum of these
three heat transfers is indeed zero, as required by the conservation of energy (Eq. 18-22).
T f ϭ
5339.8 cal
271.9 cal/CЊ
ϭ 19.6ЊC Ϸ 20ЊC.
Additional examples, video, and practice available at WileyPLUS