Week 12: Thermodynamics

\(\Delta\) ← if you see the word "Delta" instead of a triangle, please refresh the page! Thanks :)

NOTE: this page was updated Friday, April 17, ~12pm to add more problems :)

Overview⚓︎

I recommend reviewing the following concepts, they will probably come up in recitation:

The first law of thermodynamics
- \(\Delta U = Q - W\)
- taking \(W\) to mean work done by the gas
- keeping track of signs carefully
Internal energy and heat capacities for a monatomic ideal gas
- \(U = \tfrac32 nRT\)
- so \(\Delta U = \tfrac32 nR\Delta T\)
- \(C_V = \tfrac32 R\) and \(C_P = \tfrac52 R\)
- internal energy depends only on temperature
Special thermodynamic processes
- isothermal: \(\Delta T = 0\), so \(\Delta U = 0\)
- constant volume: \(W = 0\)
- adiabatic: \(Q = 0\)
- constant pressure: \(Q = nC_P\Delta T\)
Work from a \(PV\) diagram
- area under the curve
- positive for expansion
- negative for compression
- for straight-line segments, using trapezoid areas
Using \(PV=nRT\) to compare states on a diagram
- if \(n\) is fixed, then \(T \propto PV\)
- using endpoint states to find temperature changes
- for a monatomic gas, \(\Delta U = \tfrac32 (P_fV_f - P_iV_i)\)
Heat engines and thermal efficiency
- \(e = \dfrac{W}{Q_H}\)
- \(W = Q_H - Q_C\)
- power is work per time
- keeping track of whether quantities are per cycle or per second
Carnot efficiency
- \(e_{\mathrm{Carnot}} = 1 - \dfrac{T_C}{T_H}\)
- reservoir temperatures must be in Kelvin
- real engines must have efficiency below Carnot

Practice Problems⚓︎

If you're comfortable with the "★★★" problems, you should do great during recitation.
To print these questions, simply press Ctrl + P while on this page, and it should come out formatted nicely -- just make sure to refresh first so all the math renders properly.
- For a copy without solutions, click here.

Difficulty key:

★☆☆ = beginner
★★☆ = standard
★★★ = challenging / multi-step

Thermodynamics Constants / Reminders

\(R = 8.31 \ \mathrm{J/(mol \cdot K)}\)
For a monatomic ideal gas: \(U = \tfrac32 nRT\)
\(C_V = \tfrac32 R\)
\(C_P = \tfrac52 R\)
First law: \(\Delta U = Q - W\)
Isothermal ideal-gas process: \(\Delta U = 0\)
Constant-volume process: \(W = 0\)
Adiabatic process: \(Q = 0\)
Heat-engine efficiency: \(e = \dfrac{W}{Q_H}\)
Carnot efficiency: \(e_{\mathrm{Carnot}} = 1 - \dfrac{T_C}{T_H}\)
Useful conversion: \(1\,\mathrm{kPa \cdot L} = 1\,\mathrm{J}\)
Latent heat of vaporization of water: \(L_v = 2.26 \times 10^6\,\mathrm{J/kg}\)
Power: \(1\,\mathrm{W} = 1\,\mathrm{J/s}\)

[1] First Law, Internal Energy, and Special Processes⚓︎

★☆☆
A rigid container holds \(0.200 \ \mathrm{mol}\) of a monatomic ideal gas. The gas is heated from \(310 \ \mathrm{K}\) to \(370 \ \mathrm{K}\).
- (a) Find the change in internal energy.
- (b) Find the work done by the gas.
- (c) Find the heat added to the gas.
Solution

Because the container is rigid, the volume is constant, so

\[W = 0.\]

For a monatomic ideal gas,

\[\Delta U = \frac32 nR\Delta T.\]

Here,

\[\Delta T = 370 - 310 = 60 \ \mathrm{K}.\]

So

\[\Delta U = \frac32 (0.200)(8.31)(60)=1.50 \times 10^2 \ \mathrm{J}.\]

Using the first law,

\[\Delta U = Q - W.\]

Since \(W=0\), we get

\[Q = \Delta U = 1.50 \times 10^2 \ \mathrm{J}.\]

Answer: (a) \(\Delta U = +1.50 \times 10^2 \ \mathrm{J}\); (b) \(W=0\); (c) \(Q = +1.50 \times 10^2 \ \mathrm{J}\).
★☆☆
During an isothermal expansion of an ideal gas, the gas does \(680 \ \mathrm{J}\) of work on its surroundings.
- (a) What is the change in internal energy of the gas?
- (b) How much heat flows into or out of the gas?
Solution

For an isothermal process of an ideal gas, the temperature does not change, so

\[\Delta U = 0.\]

The first law gives

\[\Delta U = Q - W.\]

Since \(\Delta U = 0\),

\[Q = W = 680 \ \mathrm{J}.\]

Because \(Q\) is positive, heat flows into the gas.

Answer: (a) \(\Delta U = 0\); (b) \(Q = +680 \ \mathrm{J}\), so heat flows into the gas.
★★☆
A sample of \(0.350 \ \mathrm{mol}\) of a monatomic ideal gas is compressed isothermally at \(300 \ \mathrm{K}\) from \(3.60 \ \mathrm{L}\) to \(1.20 \ \mathrm{L}\).
- (a) Find the change in internal energy.
- (b) Find the work done by the gas.
- (c) Find the heat flow into or out of the gas.
Solution

Since the process is isothermal for an ideal gas,

\[\Delta U = 0.\]

The work done by the gas in an isothermal process is

\[W = nRT\ln\!\left(\frac{V_f}{V_i}\right).\]

Here,

\[W = (0.350)(8.31)(300)\ln\!\left(\frac{1.20}{3.60}\right).\]

Since \(\ln(1/3)\) is negative,

\[W = -9.59 \times 10^2 \ \mathrm{J}.\]

Using the first law,

\[0 = Q - W \qquad \Rightarrow \qquad Q = W.\]

So

\[Q = -9.59 \times 10^2 \ \mathrm{J}.\]

The negative sign means heat flows out of the gas.

Answer: (a) \(\Delta U = 0\); (b) \(W = -9.59 \times 10^2 \ \mathrm{J}\); (c) \(Q = -9.59 \times 10^2 \ \mathrm{J}\), so heat flows out of the gas.
★★☆
A closed rigid container holds \(0.240 \ \mathrm{mol}\) of a monatomic ideal gas at \(290 \ \mathrm{K}\). The gas is heated until its pressure is \(2.50\) times its initial value.
- (a) What is the final temperature?
- (b) What is the change in internal energy?
- (c) How much heat is added to the gas?
Solution

Because the volume is constant,

\[\frac{P}{T} = \text{constant}.\]

So

\[T_f = T_i\left(\frac{P_f}{P_i}\right)=290(2.50)=725 \ \mathrm{K}.\]

Therefore,

\[\Delta T = 725 - 290 = 435 \ \mathrm{K}.\]

The change in internal energy is

\[\Delta U = \frac32 nR\Delta T\]

\[\Delta U = \frac32 (0.240)(8.31)(435)=1.30 \times 10^3 \ \mathrm{J}.\]

Since the volume is constant, \(W=0\), so from the first law,

\[Q = \Delta U = 1.30 \times 10^3 \ \mathrm{J}.\]

Answer: (a) \(T_f = 725 \ \mathrm{K}\); (b) \(\Delta U = +1.30 \times 10^3 \ \mathrm{J}\); (c) \(Q = +1.30 \times 10^3 \ \mathrm{J}\).
★★★
In an adiabatic process, the temperature of \(3.20 \ \mathrm{mol}\) of a monatomic ideal gas drops from \(500^\circ\mathrm{C}\) to \(140^\circ\mathrm{C}\).
- (a) How much heat is exchanged with the surroundings?
- (b) What work does the gas do?
- (c) What is the change in internal energy of the gas?
Solution

For an adiabatic process,

\[Q=0.\]

Convert the temperatures to Kelvin:

\[T_i = 500 + 273 = 773 \ \mathrm{K}, \qquad T_f = 140 + 273 = 413 \ \mathrm{K}.\]

So

\[\Delta T = 413 - 773 = -360 \ \mathrm{K}.\]

The change in internal energy is

\[\Delta U = \frac32 nR\Delta T\]

\[\Delta U = \frac32 (3.20)(8.31)(-360) = -1.44 \times 10^4 \ \mathrm{J}.\]

Using the first law,

\[\Delta U = Q - W = -W,\]

because \(Q=0\). Therefore,

\[W = -\Delta U = +1.44 \times 10^4 \ \mathrm{J}.\]

Answer: (a) \(Q=0\); (b) \(W = +1.44 \times 10^4 \ \mathrm{J}\); (c) \(\Delta U = -1.44 \times 10^4 \ \mathrm{J}\).
★★★
The temperature of a \(2.80 \ \mathrm{mol}\) sample of a monatomic ideal gas is initially \(320 \ \mathrm{K}\). Its internal energy is doubled by the addition of heat.
- (a) How much heat is needed if the gas is heated at constant volume?
- (b) How much heat is needed if the gas is heated at constant pressure?
Solution

For a monatomic ideal gas,

\[U = \frac32 nRT.\]

Since internal energy is proportional to temperature, doubling \(U\) means doubling \(T\).

So

\[T_f = 2T_i = 640 \ \mathrm{K}, \qquad \Delta T = 640 - 320 = 320 \ \mathrm{K}.\]

At constant volume,

\[Q_V = nC_V\Delta T = n\left(\frac32 R\right)\Delta T.\]

Thus,

\[Q_V = (2.80)\left(\frac32\right)(8.31)(320)=1.12 \times 10^4 \ \mathrm{J}.\]

At constant pressure,

\[Q_P = nC_P\Delta T = n\left(\frac52 R\right)\Delta T.\]

So

\[Q_P = (2.80)\left(\frac52\right)(8.31)(320)=1.86 \times 10^4 \ \mathrm{J}.\]

Answer: (a) \(Q_V = 1.12 \times 10^4 \ \mathrm{J}\); (b) \(Q_P = 1.86 \times 10^4 \ \mathrm{J}\).

[2] \(PV\) Diagrams, Work, and Heat⚓︎

★☆☆
A gas expands at constant pressure \(P = 180 \ \mathrm{kPa}\) from \(1.40 \ \mathrm{L}\) to \(4.10 \ \mathrm{L}\). What is the work done by the gas?

Solution

For a constant-pressure process,

\[W = P\Delta V.\]

Here,

\[\Delta V = 4.10 - 1.40 = 2.70 \ \mathrm{L}.\]

Therefore,

\[W = (180 \ \mathrm{kPa})(2.70 \ \mathrm{L})=486 \ \mathrm{J}.\]

Answer: \(W = +486 \ \mathrm{J}\).
★☆☆
A fixed amount of an ideal gas goes from state \(A\) to state \(B\).

\[A: (P=150 \ \mathrm{kPa},\ V=2.00 \ \mathrm{L},\ T=300 \ \mathrm{K})\]

\[B: (P=250 \ \mathrm{kPa},\ V=3.00 \ \mathrm{L})\]

What is the temperature at point \(B\)?

Solution

For a fixed amount of ideal gas,

\[\frac{PV}{T} = \text{constant}.\]

So

\[\frac{P_AV_A}{T_A} = \frac{P_BV_B}{T_B}.\]

Solving for \(T_B\),

\[T_B = T_A\left(\frac{P_BV_B}{P_AV_A}\right).\]

Substituting,

\[T_B = 300\left(\frac{(250)(3.00)}{(150)(2.00)}\right)=300(2.50)=750 \ \mathrm{K}.\]

Answer: \(T_B = 750 \ \mathrm{K}\).
★★☆
A monatomic ideal gas goes from state \(A\) to state \(C\) in two steps on a \(PV\) diagram:
- \(A \to B\): constant volume from \((1.50 \ \mathrm{L},\ 200 \ \mathrm{kPa})\) to \((1.50 \ \mathrm{L},\ 350 \ \mathrm{kPa})\)
- \(B \to C\): constant pressure from \((1.50 \ \mathrm{L},\ 350 \ \mathrm{kPa})\) to \((4.50 \ \mathrm{L},\ 350 \ \mathrm{kPa})\)
If the temperature at point \(A\) is \(200 \ \mathrm{K}\), find:
- (a) the temperature at point \(C\)
- (b) the total work done by the gas
- (c) the change in internal energy
- (d) the total heat added to the gas
Solution

Since \(n\) is fixed,

\[T \propto PV.\]

Therefore,

\[\frac{T_C}{T_A} = \frac{P_CV_C}{P_AV_A} = \frac{(350)(4.50)}{(200)(1.50)} = 5.25.\]

So

\[T_C = 5.25(200)=1050 \ \mathrm{K}.\]

For the work:
- Along \(A \to B\), \(\Delta V=0\), so \(W_{AB}=0\).
- Along \(B \to C\),
\[W_{BC} = P\Delta V = (350)(4.50-1.50)=1050 \ \mathrm{J}.\]

Thus,

\[W_{\text{tot}} = 1050 \ \mathrm{J}.\]

For a monatomic ideal gas,

\[\Delta U = \frac32(P_CV_C - P_AV_A).\]

So

\[\Delta U = \frac32[(350)(4.50) - (200)(1.50)]\]

\[\Delta U = \frac32(1575 - 300)=1.91 \times 10^3 \ \mathrm{J}.\]

Finally,

\[Q = \Delta U + W = 1912.5 + 1050 = 2.96 \times 10^3 \ \mathrm{J}.\]

Answer: (a) \(T_C = 1050 \ \mathrm{K}\); (b) \(W = +1050 \ \mathrm{J}\); (c) \(\Delta U = +1.91 \times 10^3 \ \mathrm{J}\); (d) \(Q = +2.96 \times 10^3 \ \mathrm{J}\).
★★☆
A monatomic ideal gas moves from state \(A\) to state \(B\) along a straight-line path on a \(PV\) diagram.

\[A: (V=2.00 \ \mathrm{L},\ P=120 \ \mathrm{kPa},\ T=280 \ \mathrm{K})\]

\[B: (V=5.00 \ \mathrm{L},\ P=300 \ \mathrm{kPa})\]

Find:
- (a) the temperature at point \(B\)
- (b) the work done by the gas
- (c) the change in internal energy
- (d) the heat added to the gas
Solution

For the endpoint temperatures,

\[\frac{T_B}{T_A} = \frac{P_BV_B}{P_AV_A} = \frac{(300)(5.00)}{(120)(2.00)} = 6.25.\]

So

\[T_B = 6.25(280)=1750 \ \mathrm{K}.\]

Because the path is a straight line, the work is the area of a trapezoid:

\[W = \frac{P_A + P_B}{2}(V_B - V_A).\]

Thus,

\[W = \frac{120 + 300}{2}(5.00 - 2.00)=210(3.00)=630 \ \mathrm{J}.\]

The change in internal energy is

\[\Delta U = \frac32(P_BV_B - P_AV_A)\]

\[\Delta U = \frac32[(300)(5.00) - (120)(2.00)]\]

\[\Delta U = \frac32(1500 - 240)=1890 \ \mathrm{J}.\]

Then

\[Q = \Delta U + W = 1890 + 630 = 2520 \ \mathrm{J}.\]

Answer: (a) \(T_B = 1750 \ \mathrm{K}\); (b) \(W = +630 \ \mathrm{J}\); (c) \(\Delta U = +1890 \ \mathrm{J}\); (d) \(Q = +2520 \ \mathrm{J}\).
★★★
A monatomic ideal gas moves along the following four-segment path on a \(PV\) diagram:
- from \(A=(2.00 \ \mathrm{L},\ 100 \ \mathrm{kPa})\) horizontally to \((3.00 \ \mathrm{L},\ 100 \ \mathrm{kPa})\)
- then linearly to \((5.00 \ \mathrm{L},\ 250 \ \mathrm{kPa})\)
- then horizontally to \((7.00 \ \mathrm{L},\ 250 \ \mathrm{kPa})\)
- then linearly to \(B=(9.00 \ \mathrm{L},\ 150 \ \mathrm{kPa})\)
If the temperature at point \(A\) is \(240 \ \mathrm{K}\), find:
- (a) the temperature at point \(B\)
- (b) the total work done by the gas
- (c) the change in internal energy
- (d) the total heat added to the gas
Solution

First use the endpoint states to find the temperature at \(B\):

\[\frac{T_B}{T_A} = \frac{P_BV_B}{P_AV_A} = \frac{(150)(9.00)}{(100)(2.00)} = 6.75.\]

So

\[T_B = 6.75(240)=1620 \ \mathrm{K}.\]

Now compute the work by adding the areas of the four segments.

For the first horizontal segment,

\[W_1 = (100)(3.00-2.00)=100 \ \mathrm{J}.\]

For the second straight-line segment,

\[W_2 = \frac{100+250}{2}(5.00-3.00)=175(2.00)=350 \ \mathrm{J}.\]

For the third horizontal segment,

\[W_3 = (250)(7.00-5.00)=500 \ \mathrm{J}.\]

For the fourth straight-line segment,

\[W_4 = \frac{250+150}{2}(9.00-7.00)=200(2.00)=400 \ \mathrm{J}.\]

Therefore,

\[W_{\text{tot}} = 100 + 350 + 500 + 400 = 1350 \ \mathrm{J}.\]

The change in internal energy is

\[\Delta U = \frac32(P_BV_B - P_AV_A)\]

\[\Delta U = \frac32[(150)(9.00) - (100)(2.00)]\]

\[\Delta U = \frac32(1350 - 200)=1725 \ \mathrm{J}.\]

Finally,

\[Q = \Delta U + W = 1725 + 1350 = 3075 \ \mathrm{J}.\]

Answer: (a) \(T_B = 1620 \ \mathrm{K}\); (b) \(W = +1350 \ \mathrm{J}\); (c) \(\Delta U = +1725 \ \mathrm{J}\); (d) \(Q = +3075 \ \mathrm{J}\).
★★★
A monatomic ideal gas is taken from state \(A\) to state \(C\), where

\[A: (V=2.00 \ \mathrm{L},\ P=150 \ \mathrm{kPa},\ T=300 \ \mathrm{K})\]

\[C: (V=6.00 \ \mathrm{L},\ P=300 \ \mathrm{kPa}).\]

Compare the following two paths between the same endpoints:
- Path 1: a straight-line path from \(A\) directly to \(C\)
- Path 2: first go vertically to \((2.00 \ \mathrm{L},\ 300 \ \mathrm{kPa})\), then horizontally to \(C\)
For each path, find:
- (a) the work done by the gas
- (b) the heat added to the gas
Also find:
- (c) the change in internal energy between \(A\) and \(C\)
- (d) which path requires more heat input
Solution

Since the endpoints are the same, the change in internal energy is the same for both paths.

\[\Delta U = \frac32(P_CV_C - P_AV_A).\]

So

\[\Delta U = \frac32[(300)(6.00) - (150)(2.00)]\]

\[\Delta U = \frac32(1800 - 300)=2250 \ \mathrm{J}.\]

For Path 1, the work is the area of a trapezoid:

\[W_1 = \frac{150+300}{2}(6.00-2.00)=225(4.00)=900 \ \mathrm{J}.\]

Therefore,

\[Q_1 = \Delta U + W_1 = 2250 + 900 = 3150 \ \mathrm{J}.\]

For Path 2:
- the vertical segment does no work
- the horizontal segment gives
\[W_2 = (300)(6.00-2.00)=1200 \ \mathrm{J}.\]

So

\[Q_2 = \Delta U + W_2 = 2250 + 1200 = 3450 \ \mathrm{J}.\]

Since \(Q_2 > Q_1\), Path 2 requires more heat input.

Answer: (a) \(W_1 = 900 \ \mathrm{J}\) and \(W_2 = 1200 \ \mathrm{J}\); (b) \(Q_1 = 3150 \ \mathrm{J}\) and \(Q_2 = 3450 \ \mathrm{J}\); (c) \(\Delta U = +2250 \ \mathrm{J}\); (d) Path 2 requires more heat input.

[3] Heat Engines and Carnot Efficiency⚓︎

★☆☆
A heat engine absorbs \(2.20 \times 10^3 \ \mathrm{J}\) of heat each cycle and exhausts \(1.50 \times 10^3 \ \mathrm{J}\) each cycle.
- (a) How much work is done per cycle?
- (b) What is the engine's efficiency?
Solution

The work per cycle is

\[W = Q_H - Q_C = 2.20 \times 10^3 - 1.50 \times 10^3 = 700 \ \mathrm{J}.\]

The efficiency is

\[e = \frac{W}{Q_H} = \frac{700}{2200}=0.318.\]

So the efficiency is

\[e = 31.8\%.\]

Answer: (a) \(W = 700 \ \mathrm{J}\); (b) \(e = 31.8\%\).
★☆☆
A heat engine does \(480 \ \mathrm{J}\) of work in each cycle, and each cycle lasts \(0.250 \ \mathrm{s}\). What is the power output of the engine?

Solution

Power is work divided by time:

\[P = \frac{W}{\Delta t} = \frac{480}{0.250}=1920 \ \mathrm{W}.\]

Answer: \(P = 1.92 \times 10^3 \ \mathrm{W}\).
★★☆
A heat engine operates at three-quarters of the efficiency of a Carnot engine working between a hot reservoir at \(525^\circ\mathrm{C}\) and a cold reservoir at \(75^\circ\mathrm{C}\).

What is the operating efficiency of the engine?

Solution

First convert the temperatures to Kelvin:

\[T_H = 525 + 273 = 798 \ \mathrm{K}, \qquad T_C = 75 + 273 = 348 \ \mathrm{K}.\]

The Carnot efficiency is

\[e_C = 1 - \frac{T_C}{T_H} = 1 - \frac{348}{798}=0.564.\]

The actual engine runs at three-quarters of this value:

\[e = 0.75e_C = 0.75(0.564)=0.423.\]

So the operating efficiency is about

\[e = 42.3\%.\]

Answer: \(e \approx 42.3\%\).
★★☆
The efficiency of a heat engine is \(55.0\%\) of the efficiency of a Carnot engine operating between \(45^\circ\mathrm{C}\) and \(245^\circ\mathrm{C}\). If the engine absorbs heat at a rate of \(20.0 \ \mathrm{kW}\), at what rate is heat exhausted?

Solution

Convert the temperatures to Kelvin:

\[T_C = 45 + 273 = 318 \ \mathrm{K}, \qquad T_H = 245 + 273 = 518 \ \mathrm{K}.\]

The Carnot efficiency is

\[e_C = 1 - \frac{318}{518}=0.386.\]

The actual efficiency is

\[e = 0.550(0.386)=0.212.\]

For a heat engine,

\[e = 1 - \frac{\dot Q_C}{\dot Q_H}.\]

So

\[\frac{\dot Q_C}{\dot Q_H} = 1-e = 1-0.212=0.788.\]

Therefore,

\[\dot Q_C = 0.788(20.0 \ \mathrm{kW}) = 15.8 \ \mathrm{kW}.\]

Answer: The heat is exhausted at a rate of \(15.8 \ \mathrm{kW}\).
★★★
A heat engine has a thermal efficiency of \(34.0\%\). Its heat input each cycle is supplied by the condensation of \(4.20 \ \mathrm{kg}\) of steam at \(100^\circ\mathrm{C}\).
- (a) How much heat is absorbed each cycle?
- (b) How much work is done each cycle?
- (c) How much heat is lost to the surroundings each cycle?
Solution

The heat input comes from condensation:

\[Q_H = mL_v = (4.20)(2.26 \times 10^6)=9.49 \times 10^6 \ \mathrm{J}.\]

The efficiency is

\[e = \frac{W}{Q_H},\]

so the work done is

\[W = eQ_H = (0.340)(9.49 \times 10^6)=3.23 \times 10^6 \ \mathrm{J}.\]

The heat exhausted is

\[Q_C = Q_H - W = 9.49 \times 10^6 - 3.23 \times 10^6 = 6.26 \times 10^6 \ \mathrm{J}.\]

Answer: (a) \(Q_H = 9.49 \times 10^6 \ \mathrm{J}\); (b) \(W = 3.23 \times 10^6 \ \mathrm{J}\); (c) \(Q_C = 6.26 \times 10^6 \ \mathrm{J}\).
★★★
A proposed ocean-thermal power plant would operate between warm surface water at \(24^\circ\mathrm{C}\) and deep water at \(7^\circ\mathrm{C}\).
- (a) What is the maximum possible efficiency of such a plant?
- (b) If the electrical power output is \(80.0 \ \mathrm{MW}\), how much heat must be absorbed from the warm reservoir each hour, assuming the plant operates at that maximum efficiency?
Solution

The maximum possible efficiency is the Carnot efficiency.

Convert the temperatures to Kelvin:

\[T_H = 24 + 273 = 297 \ \mathrm{K}, \qquad T_C = 7 + 273 = 280 \ \mathrm{K}.\]

So

\[e_C = 1 - \frac{T_C}{T_H} = 1 - \frac{280}{297}=0.0572.\]

Thus the maximum possible efficiency is

\[e_C = 5.72\%.\]

Next, convert the power output into work per hour:

\[W = Pt = (80.0 \times 10^6)(3600)=2.88 \times 10^{11} \ \mathrm{J}.\]

Since

\[e = \frac{W}{Q_H},\]

the required heat input is

\[Q_H = \frac{W}{e_C} = \frac{2.88 \times 10^{11}}{0.0572}=5.03 \times 10^{12} \ \mathrm{J}.\]

Answer: (a) \(e_{\max} = 5.72\%\); (b) \(Q_H \approx 5.03 \times 10^{12} \ \mathrm{J}\) each hour.