Home Page Textbooks Physics Page 406: Lesson Check

Physics

1st Edition

Walker

ISBN: 9780133256925

Textbook solutions

All Solutions

Page 406: Lesson Check

Exercise 33

Step 1

1 of 2

This is the statement of the second law of thermodynamics.

Result

2 of 2

Click here for the solution.

Exercise 34

Step 1

1 of 2

It says that this is impossible since it is impossible in a cyclic process to convert all of the absorbed heat to work.

Result

2 of 2

Click here for the answer.

Exercise 35

Step 1

1 of 2

It does not because you have to put in work in order to freeze the water which increases entropy elsewhere at least for the amount it decreased when ice formed from liquid water.

Result

2 of 2

Click here for the answer.

Exercise 36

Step 1

1 of 2

The increase in entropy increases the randomness of the system.

Result

2 of 2

Click here for the solution.

Exercise 37

Step 1

1 of 2

We here always decide in favor of the system that has less order in it

a) Resulting popcorn

b) Omelet

c) a pile of bricks

d) ash.

Result

2 of 2

Click here for the solution.

Exercise 38

Step 1

1 of 2

The efficiency is calculated using $eta=1-T_{c}/T_{h}$ which gives the required order:

B($4.5%$),D($17.7%$),C($25%$),A($50%$).

Result

2 of 2

Click here for the solution.

Exercise 39

Step 1

1 of 2

The maximum efficiency is that of a Carnot’s engine working on these reservoirs and it is

$$
eta = 1-frac{T_c}{T_h} = 21%.
$$

Result

2 of 2

Click here for the solution.

Exercise 40

Step 1

1 of 2

This is the Carnot’s engine and its’ efficiency is calculated using

$$
eta = 1-frac{T_c}{T_h}
$$
which yields

$$
T_c = (1-eta)T_h = 311.6text{ K}.
$$

Result

2 of 2

Click here for the solution.

Exercise 41

Step 1

1 of 2

We can calculate this using the following sequence

$$
eta = frac{W}{Q_{received}} = frac{W}{Q_{exhausted} + W} = 28.1%.
$$

Result

2 of 2

Click here for the solution.

Exercise 42

Step 1

1 of 2

a) We know that for ideal heat engine
$$
frac{Q_c}{Q_h} = frac{T_c}{T_h}.
$$
Knowing that
$$
Q_h=W+Q_c
$$
we get

$$
frac{Q_c}{Q_c+W}=frac{T_c}{T_h}.
$$
Last two equations give us

$$
Q_c = Wfrac{T_c}{T_h-T_c} =1920text{ J}quad Q_h = 3120text{ J}
$$

so we need to add $Q_h = 3120text{ J}$

b) The heat discarded is simply
$$
Q_c=1920text{ J}.
$$

Result

2 of 2

Click here for the solution.

Exercise 43

Step 1

1 of 2

The change in entropy $Delta S$ is defined as the heat divided by the temperature,

$$
begin{align*}
Delta S &= dfrac{text{heat}}{text{temperature}}\
\
Delta S &= dfrac{Q}{T}\
end{align*}
$$

First we need to find the heat:

$$
Q = m cdot L_f
$$

Where $L_f = -33.5 cdot 10^4 text{ }dfrac{text{J}}{text{kg}}$

And $m = 3.1 text{ kg}$

Therefore,

$$
begin{align*}
Q &= 3.1 cdot (-33.5 cdot 10^4)\
\
Q &= -103.85 cdot 10^4 text{ J}\
end{align*}
$$

Temperature is $T = 0^circtext{C}$ which can be written as $T = 273text{ K}$
The change in entropy will be:

$$
begin{align*}
Delta S &= dfrac{-103.85 cdot 10^4}{273}\
\
Delta S &= -3.8 cdot 10^3 text{ J}\
\
Delta S &= boxed{-3.8 text{ kJ}}\
end{align*}
$$

Result

2 of 2

$$
Delta S = -3.8 text{ kJ}
$$

unlock

All Solutions

Page 406: Lesson Check

Haven't found what you were looking for?

Search for samples, answers to your questions and flashcards