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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 392: Lesson Check

Exercise 7

Step 1

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It extends it by recognizing so called thermal energy and heat and says that the sum of the change in the thermal energy and work done by the system equals the total amount of transfered heat to the system.

Result

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Exercise 8

Step 1

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The energy first increases i.e. the engine receives heat and then it decreases i.e. the engine releases heat (and does some work in both stages). The energy of the engine at the end of each cycle always remains the same.

Result

2 of 2

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Exercise 9

Step 1

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From the definition of efficiency ($eta = W/Q_{received}$) we see that its’ increase is indicated in Case B.

Result

2 of 2

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Exercise 10

Step 1

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The kinetic energy lost due to friction is transformed mostly in heat (and some of it in deforming the body) and this does not violate the conservation of energy.

Result

2 of 2

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Exercise 11

Step 1

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The respective changes are:
$$begin{tabular}{|c|c|c|c|c|}
hline
text{System} & textbf{A} &textbf{B}&textbf{C}&textbf{D}\
hline
W(text{J}) & 10 & -10 &30 & -20\
hline
Q(text{J})&20&-20&-50&-10\
hline
U(text{J})&10&-10&-80&10\
hline
end{tabular}
$$
If we look at the absolute change in thermal energy we see that A,B and D are tied at the change of $10text{ J}$ while the absolute change in C is $80text{ J}$ and is the highest.

Result

2 of 2

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Exercise 12

Step 1

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If we calculate the efficiency by its’ definition $eta=W/Q_h=(Q_h-Q_c)/Q_h=1-frac{Q_c}{Q_h}$ we get
$$begin{tabular}{|c|c|c|c|c|}
hline
text{System} & textbf{A} &textbf{B}&textbf{C}&textbf{D}\
hline
$Q_h$(text{J}) & 40 & 140 &80 & 240\
hline
$Q_c$(text{J})&20&120&40&220\
hline
$eta$($%$)&50.0&14.3&50.0&8.3\
hline
end{tabular}
$$
Ranking engines by increasing efficiency we get D, B, tie(C,A).

Result

2 of 2

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Exercise 13

Step 1

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Directly from the 1st law of thermodynamics we calculate
$$
Q=Delta U + W = -20text{ J} + 10text{ J} = -10text{ J}.
$$

Result

2 of 2

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Exercise 14

Step 1

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From the first law of thermodynamics we have

$$
Q=Delta U + W = -40text{ J} +100text{ J} = 60text{ J}.
$$

Result

2 of 2

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Exercise 15

Step 1

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Obviously $W=6.7times 10^5text{ J}$ and $Q= -4.1times 10^5text{ J}$. From the 1st law of thermodynamics we get

$$
Delta E = Q – W = – 10.8times 10^5text{ J}.
$$

Result

2 of 2

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Exercise 16

Step 1

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### Knowns

– Amount of heat received from the hot reservoir $Q_{text{h}} = 1220text{ J}$

– Amount of heat lost to cold reservoir $Q_{text{c}} = 680text{ J}$

Step 2

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section*{Calculation}
begin{enumerate}[a)]
item
The energy that goes into a heat engine also goes out so we write:
begin{align*}
Q_{text{h}} = Q_{text{c}} + W
end{align*}
Rearranging for work we have:
begin{align*}
W = Q_{text{h}} – Q_{text{c}}
end{align*}
Plugging in the numbers we get:
begin{align*}
W = 1220text{ J} – 680text{ J} = 540text{ J}
end{align*}
item
Now we use the definition of efficiency to find its value:
begin{align*}
e = frac{W}{Q_{text{h}}} = frac{Q_{text{h}} – Q_{text{c}} }{Q_{text{h}}} = 1 – frac{Q_{text{c}}}{Q_{text{h}}}
end{align*}
Plugging in the numbers:
begin{align*}
e = 1 – frac{680text{ J}}{1220text{ J}} approx 44.26 , %
end{align*}
end{enumerate}

Result

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begin{enumerate}[a)]
item
the work done by the engine is $W = 540text{ J}$
item
the efficiency of the engine is $e = 44.26 , %$
end{enumerate}

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