According to the 1st law of thermodynamics we have

$$
Q=Delta E + W.
$$
Noting that
$$
Delta E = 0
$$
we see that

$$
W=Q=100text{ J}.
$$

The efficiency of our engines is calculated by definition:

$$
begin{equation*}
e = frac{W}{Q_{text{h}}}
end{equation*}
$$

Where $W$ is the amount of work done, and $Q_{text{h}}$ is the amount of heat absorbed.

For the first engine we have:

$$
begin{align*}
e_1 = frac{W_1}{Q_{text{h1}}} = frac{20text{ J}}{100text{ J}} = 20 , %
end{align*}
$$

For the second engine we have:

$$
begin{align*}
e_2 = frac{W_2}{Q_{text{h2}}} = frac{60text{ J}}{text{600text{ J}}} = 10 , %
end{align*}
$$

Obviously the efficiency of engine 1 is higher than the efficiency of engine 2 ($e_1 > e_2$)

The efficiency of engine 1 is higher than the efficiency of engine 2 ($e_1 > e_2$)

We identify $W=8.2times 10^5 text{ J}$, $Q=-3.3times 10^5text{ J}$. From the first law of thermodynamics we have

$$
Delta E = Q-W = -11.5times10^5text{ J}
$$

We here just use $Q=Delta E + W$ for each part:

a) $Q=100text{ J}$

b) $Q=-100text{ J}$

c) $Q= 200text{ J}$.

a) We calculate this by definition

$$
eta = frac{W}{Q_{received}} = 21%.
$$

b) From the first law of thermodynamics and the fact that in cyclic process $Delta E=0$ we find

$$
Q=Q_h-Q_c=W
$$
and this gives

$$
Q_c=Q_h-W =610text{ J}.
$$

This can be calculated in the following procedure

$$
eta = frac{W}{Q_{received}} = frac{Q_{received}-Q_{exhausted}}{Q_{received}} = 1-frac{Q_{exhausted}}{Q_{received}} = 23%.
$$

This can be calculated in the following procedure

$$
eta = frac{W}{Q_{h}} = frac{Q_{h}-Q_{c}}{Q_{h}} = 1-frac{Q_{c}}{Q_{h}} = 1-frac{2}{3} = 33%.
$$

### Knowns

– The amount of heat absorbed from the hot reservoir $Q_{text{h}} = 280text{ J}$

### Calculation

From the figure in the problem we know that our engine operates with such efficiency that

$$
begin{align*}
Q_{text{c}} = frac{2}{3} , Q_{text{h}}
end{align*}
$$

Knowing this, and using the first law of thermodynamics we calculate the work done by the engine:

$$
begin{align*}
W = Q_{text{h}} – Q_{text{c}} = Q_{text{h}} – frac{2}{3} , Q_{text{h}} = frac{1}{3} , Q_{text{h}}
end{align*}
$$

Plugging in the numbers we get:

$$
begin{align*}
W = frac{280text{ J}}{3} = 93.3text{ J}
end{align*}
$$

The amount of work done by the engine is $W = 93.3text{ J}$

From the first law of thermodynamics we know that

$$
Q=Delta E + WRightarrow Delta E = Q-W.
$$
If we suppose all of the heat is absorbed by sweat and used for its’ evaporation we can write $Q=-lambda_e m$ (minus sign is due to the fact that the heat is released by the player) which yields

$$
Delta E = – lambda_e m -W = – 491.6text{ kJ}.
$$

We will always use the 1st law of thermodynamics in the form of
$$
Q=Delta E + W.
$$

a) $Delta E = Q-W = 77text{ J}-(-42text{ J}) =119text{ J}$;

b)$Delta E = Q-W = 77text{ J} – 42text{ J} = 35text{ J}$;

c) $Q=-120text{ J} + 120text{ J} = 0.$

In the constant-volume process there is no work done so $W=0$.

Thermal energy of the system is determined by its’ temperature so if there is no change in temperature $Delta E = 0$.

By definition of adiabatic process there is no exchange of heat between the system and its’ surroundings so $Q=0$.

The work is determined as the area under the graph which here reduces to the area of a rectangle whose one side is $p$ (pressure) and the other side is $Delta V$ (change in volume).

We know that for a gas expanding in a constant pressure process the work can be calculated as the area beneath the curve representing the process on a pressure-volume diagram.

This curve is actually a horizontal line, and the area is that of a rectangle with sides $P$ and $Delta V$.

We write:

$$
begin{align*}
W = P cdot Delta V
end{align*}
$$

We see that as $P$ increases the amount of work will also increase.

The work is equal to the area of rectangle made by $p$ and $Delta V$:

$$
W=pDelta V =50.82text{ kJ}.
$$

The work is equal to the area of rectangle made by $p$ and $Delta V$:

$$
W=pDelta V =p(2V_i-V_i) = pV_i =92.4text{ kJ} .
$$

The work is equal to the area of rectangle made by $p$ and $Delta V$:

$$
W=pDelta V
$$
which yields
$$
Delta V=frac{W}{p} = 4.3times 10^{-4}text{ m}^3.
$$

a) Since the work done from the point of view of internal forces of the gas is negative its’ volume decreases.

b) We know that the work in isobaric process is calculated as

$$
W=pDelta V
$$
which yields

$$
Delta V = frac{W}{p} = – 5.4times 10^{-4}text{ m}^3.
$$
Note that the work done by the gas is $W=-62text{ J}$ so volume decreases (the change is negative).

The work done in isobaric process is simply
$$
W=pDelta V
$$
which gives us

$$
p=frac{W}{Delta V} = 29text{ Pa}.
$$

a) In adiabatic process, since $Q=0$ we simply get $W=-Delta E$. Since in this problem we suppose a positive change in thermal energy we conclude that the work is negative which is equivalent to work being done on the system.

b) From the equation from part a) we see that $W=-890text{ J}$ which means that $890text{ J}$ of work are done on the system.

The constant pressure means the system is isobaric. Hence, the work done by the system is:

$$
W=pDelta V=(125)cdot(0.75)=93.75:kJ
$$

The $1$st law of thermodynamics states:

$$
Q=Delta E+W
$$

Substitute the change in thermal energy $Delta E$ given in the task to find the heat flow $Q$.

$$
textbf{a.)}; Delta E=65:J:
$$

$$
Q=65+93750=93815:J=boxed{93.815:kJ}
$$

$Q$ is positive, so the heat flows into the system.

$$
textbf{b.)}; Delta E=-1850:J:
$$

$$
Q=-1850+93750=91900:J=boxed{91.9:kJ}
$$

$Q$ is positive, so the heat flows into the system.

$textbf{a.)}:Q=93.815:kJ$

$textbf{b.)}:Q=91.9:kJ$

In both cases the heat flows into the system.

We know that the maximum efficiency of a heat engine operating between the Kelvin temperatures $T_{text{h}}$ and $T_{text{c}}$ is:

$$
begin{align*}
e_{text{max}} = 1 – frac{T_{text{c}}}{T_{text{h}}}
end{align*}
$$

So we see that as $frac{T_{text{c}}}{T_{text{h}}}$ increases $e_{text{max}}$ decreases.

Now we will se how $frac{T_{text{c}}}{T_{text{h}}}$ changes when both temperatures are increased for the same amount(50 K in our problem).

We know that $T_{text{c}}$ is always smaller than $T_{text{h}}$ by definition, which means that the same absolute change in both temperatures will bring about a greater relative change in the temperature $T_{text{c}}$ than in the temperature $T_{text{h}}$.

This means that $frac{T_{text{c}}}{T_{text{h}}}$ increases when both temperatures are increased by $50text{ K}$. We have already shown that this produces a decreases in $e_{text{max}}$

The efficiency is an increasing function of the ratio $T_h/T_c$ so the order of increasing efficiency matches that of the increasing ratio $T_h/T_c$ and it is easily checked that it reads:

D, Tie (C, B), A.

By definition it is

$$
eta=frac{W}{Q_{received}} = 38%.
$$

a) Using 1st law of thermodynamics and the fact that $Delta E= 0$ in cyclic processes we have

$$
Q=WRightarrow Q_h-Q_c = W
$$
and that yields
$$
Q_c= Q_h-W = 670text{ J}.
$$

b) The efficiency by definition is

$$
eta = frac{W}{Q_h} =14%.
$$

a) Since $250text{ MW}$ is used off of $835text{ MW}$, the remainder of $585text{ MW}$ is discarded to the environment.

b) The efficiency is simply the ratio of utilized power and the invested power which is here

$$
eta = frac{P_u}{P_i} = 30%.
$$

a) The output is the invested power multiplied by efficiency which gives
$P_u =eta P_i$. What is discarded to the environment is the difference between the invested power and the output which guides us to
$$
P_{discarded} = P_i-P_u = P_i-eta P_i = P_i(1-eta) = P_ufrac{1-eta}{eta} approx 1170text{ MW}.
$$

b) We already explained
$$
P_i = frac{P_u}{eta} =1720text{ MW}.
$$

Since the water is already at boiling temperature, this temperature remains constant as it evaporates so the change in entropy is simply

$$
Delta S = frac{Q}{T_b} = frac{lambda_b m}{T_b} = 11.2text{ kJ/K}.
$$

Note that we used the boiling temperature in kelvins $T_b=373text{ K}.$

By definition of efficiency we have
$$
eta = frac{W}{Q_h} = frac{Q_h-Q_c}{Q_h} = 1-frac{Q_c}{Q_h}.
$$

a) We find this as

$$
Q_h = frac{W}{eta} =15000text{ J}.
$$

b) We find this as
$$
Q_c = Q_h-W = 12300text{ J}.
$$

c) It decreases since then we will need less heat for the same amount of work.

### Knowns

– The efficiency of a heat engine $e_1 = 21 , %$

– The new efficiency of a heat engine $e_2 = 25 , %$

– The temperature of the cold reservoir $T_{text{c1}} = 265text{ K}$

### Calculation

To find the efficiency of a heat engine operating with hot and cold reservoirs at temperatures $T_{text{h}}$ and $T_{text{c}}$ we use the formula:

$$
begin{equation*}
e = 1 – frac{T_{text{c}}}{T_{text{h}}}
end{equation*}
$$

we will apply this formula for old and new engine. We get:

$$
begin{align*}
e_1 = 1 – frac{T_{text{c1}}}{T_{text{h}}} qquad e_2 = 1 – frac{T_{text{c2}}}{T_{text{h}}}
end{align*}
$$

Where we have used the fact that the hot reservoir temperature is constant.

We will now solve the system for $T_{text{c2}}$

$$
begin{align*}
T_{text{h}} &= frac{T_{text{c1}}}{1 – e_1} = frac{T_{text{c2}}}{1 – e_2} \\
T_{text{c}} &= frac{1 – e_2}{1- e_1} , T_{text{c1}}
end{align*}
$$

Plugging in the numbers we get:

$$
begin{align*}
T_{text{c2}} = frac{1 – 0.25}{1 – 0.21} cdot 265text{ K} approx 251.6text{ K}
end{align*}
$$

The required cold reservoir temperature is $T_{text{c}} = 251.6text{ K}$

a) We can determine this from $eta = 1-frac{T_c}{T_h}$ and that yields

$$
T_c = (1-eta)T_h =381.5text{ K}.
$$

b) From the equation for calculating the efficiency it is clear that we need to decrease the temperature of the low temperature reservoir in order to increase its’ efficiency.

c) In completely identical procedure used in part a) we determine

$$
T’_c = (1-eta’)T_h = 327text{ K}.
$$

a) By its’ definition we determine
$$
eta = frac{W}{Q_h} = 88%.
$$

b) We determine this from
$$
W=Q_h-Q_cRightarrow Q_c=Q_h-W = 300text{ J}.
$$

c) If we assume an ideal engine we find

$$
frac{T_h}{T_c}=frac{Q_h}{Q_c} =8.33.
$$

### Knowns

– The rate of energy leaking $P = 25text{ kW}$

– The inside temperature $T_{text{i}} = 22text{textdegree}text{C}$

– The outside temperature $T_{text{o}} = -14text{textdegree}text{C}$

### Calculation

The total change in entropy of the universe is the sum of the decrease inside the house, and the increase outside the house. The rate of entropy increase is found as follows:

$$
begin{align*}
frac{Delta S}{Delta t} &= frac{Delta S_1}{Delta t} + frac{Delta S_2}{Delta t} \\
frac{Delta S}{Delta t} &= frac{Q_1}{T_1 , Delta t} + frac{Q_2}{T_2 , Delta t}
end{align*}
$$

The index 1 for the general case will become “i” and the index 2 will become “o”.

So we write:

$$
begin{align*}
Q_1 = Q_{text{i}} quad,quad Q_2 = Q_{text{o}} quad,quad T_1 = T_{text{i}} quad,quad T_2 = T_{text{o}}
end{align*}
$$

Since heat leaves from the inside to the outside at the rate $P$, we can write:

$$
begin{align*}
Q_{text{i}} &= – P , Delta t \
Q_{text{o}} &= P , Delta t
end{align*}
$$

Now the rate of entropy change becomes:

$$
begin{align*}
frac{Delta S}{Delta t} = frac{- P , Delta t}{T_{text{i}} , Delta t} + frac{P , Delta t}{T_{text{o}} , Delta t}
= P left(- frac{1}{T_{text{i}}} + frac{1}{T_{text{o}}} right)
end{align*}
$$

Plugging in the numbers we get:

$$
begin{align*}
frac{Delta S}{Delta t} &= 25text{ kW} left(- frac{1}{(22 + 273.15)K} + frac{1}{(-14 + 273.15)K}right) \
frac{Delta S}{Delta t} &= – 84.7 , frac{text{W}}{text{K}} + 96.47 , frac{text{W}}{text{K}} = 11.77 , frac{text{W}}{text{K}}
end{align*}
$$

The rate of entropy increase if the universe: $frac{Delta S}{Delta t} = 11.77 , frac{text{W}}{text{K}}$

Since the kinetic energy of the parachutist is constant all of the work done by gravitational force is spent on overcoming the air resistance. It is reasonable to suppose that all of this work is converted to heat that is absorbed by the air. Knowing that the work of gravitational force can be written as the difference of the potential energy we can write

$$
Q=W_{g} = mgDelta h
$$

which finally gives
$$
Delta S = frac{Q}{T} = frac{mgDelta h}{T} = 1057text{ J/K}.
$$

The work done is equal to the area under the graph. This reduces to calculating the area of the shaded right trapezoid in the picture. Identifying its’ bases with $p_A =150text{ kPa}$ and $p_B = 50text{ kPa}$ and its’ height with $Delta V = 4text{ m}^3 – 1text{ m}^3 = 3text{ m}^3$ and knowing basic geometry we get

$$
W=frac{1}{2}(p_A+p_B)Delta V = 300000text{ J}.
$$

The maximum efficiency is that of Carnot’s engine given by

$$
eta = 1-frac{T_c}{T_h} = 6.1%.
$$

(Temperatures are converted to kelvins)
which corresponds to A.

This is calculated from the formula

$$
Q=mc_w(T_2-T_1) = -1.1times 10^8text{ J}
$$
which points to B.

Similarly like in the problem 92. we find

$$
eta = 1-frac{T_c}{T_h} = 6.8%
$$

which points to A.