All Solutions
Page 410: Assessment
$$
Q=Delta E + W.
$$
Noting that
$$
Delta E = 0
$$
we see that
$$
W=Q=100text{ J}.
$$
$$
begin{equation*}
e = frac{W}{Q_{text{h}}}
end{equation*}
$$
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 ($e_1 > e_2$)
$$
Delta E = Q-W = -11.5times10^5text{ J}
$$
a) $Q=100text{ J}$
b) $Q=-100text{ J}$
c) $Q= 200text{ J}$.
$$
eta = frac{W}{Q_{received}} = 21%.
$$
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
$$
Q_c=Q_h-W =610text{ J}.
$$
$$
eta = frac{W}{Q_{received}} = frac{Q_{received}-Q_{exhausted}}{Q_{received}} = 1-frac{Q_{exhausted}}{Q_{received}} = 23%.
$$
$$
eta = frac{W}{Q_{h}} = frac{Q_{h}-Q_{c}}{Q_{h}} = 1-frac{Q_{c}}{Q_{h}} = 1-frac{2}{3} = 33%.
$$
– The amount of heat absorbed from the hot reservoir $Q_{text{h}} = 280text{ J}$
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:
$$
begin{align*}
W = frac{280text{ J}}{3} = 93.3text{ J}
end{align*}
$$
$$
Q=Delta E + WRightarrow Delta E = Q-W.
$$
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
$$
Delta E = – lambda_e m -W = – 491.6text{ kJ}.
$$
$$
Q=Delta E + W.
$$
a) $Delta E = Q-W = 77text{ J}-(-42text{ J}) =119text{ J}$;
b)$Delta E = Q-W = 77text{ J} – 42text{ J} = 35text{ J}$;
c) $Q=-120text{ J} + 120text{ J} = 0.$
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.
$$
W=pDelta V =50.82text{ kJ}.
$$
$$
W=pDelta V =p(2V_i-V_i) = pV_i =92.4text{ kJ} .
$$
$$
W=pDelta V
$$
which yields
$$
Delta V=frac{W}{p} = 4.3times 10^{-4}text{ m}^3.
$$
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).
$$
W=pDelta V
$$
which gives us
$$
p=frac{W}{Delta V} = 29text{ Pa}.
$$
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.
$$
W=pDelta V=(125)cdot(0.75)=93.75:kJ
$$
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.
$textbf{b.)}:Q=91.9:kJ$
In both cases the heat flows into the system.
$$
W=pDelta V = 93750text{ J}.
$$
Also note the validity of the 1st law of thermodynamics $Q=Delta E + W$.
a) Here we find $Q=93815text{ J},$
b) Here we find $Q=91900text{ J}.$
In both parts the heat is added to the system.
$$
begin{align*}
e_{text{max}} = 1 – frac{T_{text{c}}}{T_{text{h}}}
end{align*}
$$
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 $e_{text{max}}$
b) The only physically acceptable explanation is C. and its’ validity is obvious (almost as much as the invalidity of other two statements).
D, Tie (C, B), A.
$$
eta=frac{W}{Q_{received}} = 38%.
$$
$$
Q=WRightarrow Q_h-Q_c = W
$$
and that yields
$$
Q_c= Q_h-W = 670text{ J}.
$$
b) The efficiency by definition is
$$
eta = frac{W}{Q_h} =14%.
$$
b) The efficiency is simply the ratio of utilized power and the invested power which is here
$$
eta = frac{P_u}{P_i} = 30%.
$$
$P_u =eta P_i$. 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
$$
P_i = frac{P_u}{eta} =1720text{ MW}.
$$
$$
Delta S = frac{Q}{T_b} = frac{lambda_b m}{T_b} = 11.2text{ kJ/K}.
$$
Note that we used the boiling temperature in kelvins $T_b=373text{ K}.$
$$
eta = frac{W}{Q_h} = frac{Q_h-Q_c}{Q_h} = 1-frac{Q_c}{Q_h}.
$$
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.
– 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 $T_{text{c1}} = 265text{ K}$
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:
$$
begin{align*}
T_{text{c2}} = frac{1 – 0.25}{1 – 0.21} cdot 265text{ K} approx 251.6text{ K}
end{align*}
$$
$$
T_c = (1-eta)T_h =381.5text{ K}.
$$
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
$$
T’_c = (1-eta’)T_h = 327text{ K}.
$$
$$
eta = frac{W}{Q_h} = 88%.
$$
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
$$
frac{T_h}{T_c}=frac{Q_h}{Q_c} =8.33.
$$
– The rate of energy leaking $P = 25text{ kW}$
– The inside temperature $T_{text{i}} = 22text{textdegree}text{C}$
– The outside temperature $T_{text{o}} = -14text{textdegree}text{C}$
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:
$$
begin{align*}
frac{Delta S}{Delta t} &= 25text{ kW} left(- frac{1}{(22 + 273.15)K} + frac{1}{(-14 + 273.15)K}right) \
frac{Delta S}{Delta t} &= – 84.7 , frac{text{W}}{text{K}} + 96.47 , frac{text{W}}{text{K}} = 11.77 , frac{text{W}}{text{K}}
end{align*}
$$
$$
Q=W_{g} = mgDelta h
$$
which finally gives
$$
Delta S = frac{Q}{T} = frac{mgDelta h}{T} = 1057text{ J/K}.
$$
$$
W=frac{1}{2}(p_A+p_B)Delta V = 300000text{ J}.
$$
It is safe to say the all the potential energy is turning into heat via air resistance and a number of different friction forces inside the wheels, chains, and the remaining mechanism of the bicycle.
$$
eta = 1-frac{T_c}{T_h} = 6.1%.
$$
(Temperatures are converted to kelvins)
which corresponds to A.
$$
Q=mc_w(T_2-T_1) = -1.1times 10^8text{ J}
$$
which points to B.
$$
eta = 1-frac{T_c}{T_h} = 6.8%
$$
which points to A.
