Home Page Textbooks Physics: Principles and Problems Page 331: Section Review

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

Textbook solutions

All Solutions

Page 331: Section Review

Exercise 27

Solution 1

Solution 2

Step 1

1 of 1

The steam released its heat of vaporization into the room which was then circulated back to the boiler.

Step 1

1 of 1

In the boiler, water was often heated to high temperatures, to the point at which there was only water vapor sent trough the pipes and this means that the water vapor carried with itself a lot of thermal energy. This hot water vapor was then sent to the the radiators in the house where the radiators were heated up by the vapor and they were heating the room. As the water vapor in the radiators loses its thermal energy due to the interaction with the radiator it began to condense and turn back to water. This water was then brought back to the boiler where it could be heated up again.

Exercise 28

Step 1

1 of 7

**Given:**
– $m = 0.050 text{kg}$
– $T_1 = 80.0 degree text{C}$
– $T_2 = 100.0 degree text{C}$
– $T_3 = 110.0 degree text{C}$
– $C_text{water} = 4180 frac{text{J}}{text{kg} cdot text{K}}$
– $C_text{steam} = 2020 frac{text{J}}{text{kg} cdot text{K}}$
– $H_v = 2.26 times 10^6 frac{text{J}}{text{kg}}$

Step 2

2 of 7

Te process of heating water from $80 degree text{C}$ to $110 degree text{C}$ consists of three phases:
– the water is first heated from $T_1$ to $T_2$;
– then evaporation of water at a temperature of $T_2 = 100 degree text {C}$ occurs;
– and finally the water vapor is superheated to a temperature of $T_3$.

Now we will analyze each process separately.

Step 3

3 of 7

The heat required to raise the temperature of water from $T_1$ to $T_2$ is given by the equation below:

$$
Q_1 = mC_text{water}(T_2 – T_1)
$$
Plugging in the given values, we have:

$$begin{aligned}
Q_1 &= (0.050) cdot (4180) cdot (100.0 – 80.0)\

& = 4180 text{J}
end{aligned}$$

Step 4

4 of 7

The heat required to transform liquid water into steam can be calculated using the equation

$$
Q_2 = mH_v
$$

Plugging in the given values, we have

$$begin{aligned}
Q_2 &= (0.050) cdot (2.26 times 10^6)\

&= 1.13 times 10^5 text{J}
end{aligned}$$

Step 5

5 of 7

The heat required to raise the temperature of steam from $T_2$ to $T_3$ is given by the equation below

$$
Q_3 = mC_text{steam}(T_3 – T_2)
$$

Plugging in the given values, we have:

$$begin{aligned}
Q_3 &= (0.050) cdot (2020) cdot (110 – 100)\

& = 1010 text{J}
end{aligned}$$

Step 6

6 of 7

We can now compute the total heat required as follows:

$$begin{aligned}
Q_text{total} &= Q_1 + Q_2 + Q_3\

& = 4180 + 1.13 times 10^5 + 1010\
&=118190 text{ J}\

&= boxed{118.2 text{kJ}}
end{aligned}$$

Result

7 of 7

$$Q_text{total} = 118.2 text{ kJ}$$

Exercise 29

Solution 1

Solution 2

Step 1

1 of 7

hfill . \
textbf{Given:} \
$m = 1.0 text{kg}$ \
$C_text{mercury} = 140 frac{text{J}}{text{kg} cdot text{K}}$ \
$H_v = 3.06 times 10^5 frac{text{J}}{text{kg}}$ \
$T_i = 10.0 ^circ text{C}$ \
$T_f = 357 ^circ text{C}$

Step 2

2 of 7

hfill . \
textbf{Calculation:}\
The heat required to heat mercury from $T_i$ to $T_f$ is given by the equation

$$
Q_1 = mC_text{mercury}(T_f – T_i)
$$

Step 3

3 of 7

hfill . \
Pluggging in the given values, we have

$$
Q_1 = (1.0) cdot (140) cdot (357 – 10.0)
$$

$$
Q_1 = 48580 text{J}
$$

Step 4

4 of 7

hfill . \
The heat required to vaporize mercury is given by the equation

$$
Q_2 = mH_V
$$

Step 5

5 of 7

hfill . \
Plugging in the given values, we have

$$
Q_2 = (1.0) cdot (3.06 times 10^5)
$$

$$
Q_2 =3.06 times 10^5 text{J}
$$

Step 6

6 of 7

hfill . \
The total heat required can now be calculated as follows

$$
Q_text{total} = Q_1 + Q_2
$$

$$
Q_text{total} = 48580 + 3.06 times 10^5
$$

$$
boxed{Q_text{total} = 3.5 times 10^5 text{J}}
$$

Result

7 of 7

$$
Q_text{total} = 3.5 times 10^5 text{J}
$$

Step 1

1 of 2

$Q = m C_{Hg} Delta T + m H_{v}$

$Q = (1.0 kg) (140 J/kg. text{textdegree} C) (357 text{textdegree} C – 10.0 text{textdegree} C) + (1.0kg)(3.06 times 10^{5} J / kg )$

$Q = 3.5 times 10^{5} J$

Result

2 of 2

$textit$$text{color{#c34632}3.5 times 10^{5} J $$}$

Exercise 30

Solution 1

Solution 2

Step 1

1 of 1

Water at the top will have gravitational potential energy that dissipates when it begins to splash around. Water is “hotter” at the bottom than at the top.

Step 1

1 of 1

When we drop something from a certain height at the start an object has initial velocity equal to zero but it certainly falls down with some velocity that is different from zero. We say that the gravitational potential energy of an object has been converted into kinetic energy and that is why an object has greater velocity at the bottom. All of this works exactly the same with the waterfall. Water at the top has some kinetic energy and some gravitational potential energy and as it falls down gravitational potential energy is converted into the kinetic energy and so the temperature of the water at the bottom should be higher form the temperature of water at the top.

Exercise 31

Solution 1

Solution 2

Step 1

1 of 4

Kinetic energy of the hammer is absorber as heat or thermal energy through the lead block.

Step 2

2 of 4

Hammer kinetic energy = $dfrac{1}{2}mv^{2} = dfrac{1}{2} (320 kg) (5.0m/s)^{2} = 4.0 kJ$

Step 3

3 of 4

Change in heat energy = $Delta U = mC Delta T = (3.0 kg)(130 J/kgK)(5.0 C) = 2.0 kJ$

Step 4

4 of 4

Half of the hammer’s energy went to the lead block.

Step 1

1 of 5

**Given information:**

Mass of the hammer: $m_H=320:mathrm{kg}$

Mass of the lead block: $m_L=3: mathrm{kg}$

Velocity of the hammer: $v=5:mathrm{frac{m}{s}}$

Specific heat of lead: $c=130:mathrm{frac{J}{kg:K}}$

Change in temperature: $Delta T=5:{}^mathrm{o}C$

Step 2

2 of 5

**Approach:**

At the moment of impact some of the hammers kinetic energy is transferred as heat to the lead block and we will now calculate how much energy has been transferred.

Step 3

3 of 5

Kinetic energy of a hammer at the moment of impact is
$$begin{align*}
E_k&=frac{m_H v^2}{2}\
&=frac{320cdot 5^2}{2}\
&=4:mathrm{kJ}
end{align*}$$

Step 4

4 of 5

Change in the internal energy of an object is
$$begin{align*}
Delta U&=m_LcDelta T\
&=3cdot130cdot5\
&=1.95:mathrm{kJ}
end{align*}$$

Step 5

5 of 5

The amount of heat received at the impact is
$$
E_k-Delta U=2.05:mathrm{kJ}
$$
and that is why the temperature of the lead block got larger.

Exercise 32

Solution 1

Solution 2

Step 1

1 of 6

hfill . \
textbf{Given:} \
$h = 125.0 text{m}$ \
$C_text{water} = 4180 frac{text{J}}{text{kg} cdot text{K}}$ \
$g = 9.8 frac{text{m}}{text{s}^2}$

Step 2

2 of 6

hfill . \
textbf{Calculation:}\
Since the potential energy of the water is all converted to thermal energy, we can express this relationship using the equation

$$
PE = Q
$$

Step 3

3 of 6

hfill . \
We know that $PE = mgh$ and $Q = mC_text{water}Delta T$, thus we can express the equation above as follows

$$
mgh = mC_text{water}Delta T
$$

Step 4

4 of 6

hfill . \
Isolating $Delta T$ on one side of the equation, we have

$$
Delta T = frac{gh}{C_text{water}}
$$

Step 5

5 of 6

hfill . \
Plugging in the given values

$$
Delta T = frac{(9.8)(125.0)}{4180}
$$

$$
boxed{Delta T = 0.2931 ^circ text{C}}
$$

Result

6 of 6

$$
Delta T = 0.2931 ^circ text{C}
$$

Step 1

1 of 2

$P E_{gravity} = Q_{absorbed}$

$mgh = m C Delta T$ =$>$ $Delta T = dfrac{gh}{C}$

$Delta T = dfrac{(9.8 m/s^{2})(125.0m)}{4180 J /kg.text{textdegree} C}$

$Delta T = 0.293 text{textdegree} C$

Result

2 of 2

$textit$$text{color{#c34632}0.293 text{textdegree} C $$}$

Exercise 33

Step 1

1 of 1

The physical quantity that indicates the amount of disorder is called entropy. Thus, for processes in which entropy has a higher value, we can say that the amount of disorder is also higher.

Now consider the natural gas used to heat the house. During the combustion process of natural gas, heat is released and is transferred to the indoor air through the heating system. This heated indoor air does not have the ability to do the work nor can the heat received by the air be further utilized. On the other hand, the initial molecules of natural gas before combustion had a much greater potential to obtain useful work or utilize heat energy. Therefore, it is considered that heating a house with natural gas is a process in which there is a lack of usable energy, ie an increase in entropy. Consequently, the amount of disorder is increased.

Exercise 34

Solution 1

Solution 2

Step 1

1 of 1

It’s unlikely that the cards will return to their original order according to the second law of thermodynamics as the disorder increases every time.

Step 1

1 of 1

The second law of thermodynamics states that the entropy of an irreversible system always increases. In other sense it says that the system becomes more and more disordered with time. If we start with an ordered deck of cards it is very unlikely that with shuffling we will get back to the initial state in which all cards are ordered because there is a lot of possible configurations in which cards can be and we can see that this is very reminiscent of the second law of thermodynamics.

unlock

All Solutions

Page 331: Section Review

Haven't found what you were looking for?

Search for samples, answers to your questions and flashcards