Practice Problem Set 3 Solutions
CIV300/ENV346
Fall 2010
1) This question explores the basic energy conversions implied by the first law of
thermodynamics. The basic unit considered is one we can all hopefully visualize: it is the
energy released through a small car’s load of fuel, say burning in air 40 L of gasoline at 35
MJ/L. The TES system we will initially compare this to is 1 cubic meter of (pure) water, with
(for simplicity) an assumed density of 1000 kg per cubic meter.
(a) What speed would this cubic meter of water have to be travelling at to have the KE
equivalent of this small tank of gasoline?
(b) How high would you have to lift the cubic meter against a constant gravitational field of
9.8 m/s2?
(c) What would be the increase in temperature in the cubic meter of water to absorb all the
energy in thermal form associated with the gasoline?
(d) How much water would have to be evaporated, on a mass (kg) and on a fraction of the
cubic meter basis?
(e) If the water were in the form of ice, how much could be melted with this much energy?
(f)
How many photons (in Einsteins E) would be required of green light with a wavelength of
550 nm (note: an Einstein is a mole of photons).
(g) Water can be compressed with a bulk modulus of about 2 GPa. This compression does
work which is stored as elastic energy. How many cubic meters of water would have to
be compressed to store this much energy if the pressure change is associated with eight
kilometers of pure water (about the same depth as the deepest part of the ocean).
(h) How much of the mass would have to be converted directly to energy using nuclear
considerations, assuming suitable ones were available (i.e., using E = mc2)?
(i)
Assuming a net solar energy input with an average in specific area of 200 W/m2, how
much area would be required to absorb this much energy in one hour?
Solutions: Note this is as much as a computational exercise; it is one to help your work at
visualizing how big the various energy terms are.
Derived SI units: J = N⋅ m = W⋅ s = kg ⋅ m2/s2
Pa = N/m2 = kg/m⋅s2
Thermal properties of water (Liquid):
y Specific heat capacity: Cp = 4.2 J/g⋅K
y Latent heat of evaporation: LE = 2270 J/g (kJ/kg)
y Latent heat of melting: LM = 334 J/g (kJ/kg)
Total Energy released by burning the gasoline = 40 L × 35 MJ/L = 1400 MJ = 1.4 × 109 J
1 cubic meter of water, m = 1000 kg/m3 × 1 m3 = 1000 kg
a) Kinetic energy KE =
2 KE
1
=
mv 2 , then v =
m
2
1
2 ×1.4 ×109 ( J )
= 1673.3 (m / s)
1000(kg )
b) Potential energy PE = mgh, then
h=
PE
1.4 ×109 ( J )
=
= 1.43 ×105 (m) = 143(km)
mg 1000(kg ) × 9.8(m / s 2 )
This effectively puts our cubic meter in orbit. Alternatively one could visualize 1000 cubic
meters raised 143 m; if this were to dropped from this height, run away fast!
c) Consider the specific heat capacity of water, Q = C p mΔT , where Q is the heat absorbed.
Therefore, ΔT =
Q
1.4 × 109 ( J )
=
= 333.3( o K ) = 333.3( oC )
C p m 4.2( J / g ⋅ K ) × 1000 × 103 ( g )
However, the boiling point of water is 100 °C, suggesting that water evaporation will occur to the
cubic meter of water if all the energy is absorbed. Then, we need to reconsider this question as in
part d). Or, again, one could visual a “megalitre” (1000 cubic meters) raised in temperature by a
third of a degree Celsius.
d) Here it is assumed that density of water is 1000 kg/m3, which typically occurs at 4 °C. Based
on calculation from c), by absorbing the energy, water temperature will increase from 4 °C until
reaching 100 °C, and then evaporation will happen.
Energy needed to heat water from 4 °C to 100 °C:
Q = C p mΔT = 4.2 ×106 × 96 ≈ 4.2 ×108 ( J ) =4.0*10^8
Extra energy for evaporation at 100 °C: QE = 1.4 × 109 − 4.2 × 108 = 9.8 × 108 ( J ) =9.97*10^8
Mass of water evaporated:
mE = QE / LE = 9.8 × 108 ( J ) / 2270( J / g ) = 4.32 × 105 ( g ) = 432( kg ) =4.39*10^5(g)=439 kg,
about 44% of the water mass is evaporated, which is pretty impressive
e) If we only consider the melting of ice into water at 0 °C, then the mass of ice melted will be:
mM = Energy / LM = 1.4 × 109 ( J ) / 334( J / g ) = 4.2 × 106 ( g ) = 4200( kg )
f)
The Planck–Einstein equation is as follows:
E=
hc
λ
where h is the Planck constant (6.6*10-34J.s), c is the speed of light (3*108m/s) and λ is
wavelength (550 nm)
6 . 6 × 10 − 34 ( J .s ) × 3 × 10 8 (
E =
550 × 10 − 9 ( m )
m
)
s = 3 . 6 × 10 −19 J
by multiplying by the Avogadro constant, one can get the energy of one mole of photons (NA =
6.022* 1023 mol−1)
E = 3 . 6 × 10 − 19 ( J ) × 6 . 022 × 10 23 ( mol
−1
) = 216792
2
J
mol
The number of photons =
1 . 4 × 10 9 ( J )
= 6457 . 8 mol (or Einsteins)
J
216792 (
)
mol
g) Bulk modulus elasticity of water E = 2 GPa = 2 × 109 Pa = 2 × 109 kg/m⋅s2
The bulk modulus elasticity is material property characterizing the compressibility of a fluid –
how easy a unit of the fluid volume can be changed when changing the pressure working upon it.
It can be expressed as: E =
dP
dP
=
dV / V d ρ / ρ
Recall that the pressure change associated with water depth (8 km) can be expressed as:
dP = ρ gh = 103 (kg / m3 ) × 9.8(m / s 2 ) × 8000(m) = 7.84 × 107 (kg / m ⋅ s 2 ) = 7.84 × 107 ( Pa )
dP 7.84 ×107 ( Pa)
=
= 0.0392 = 3.92% , the water volume is compressed
Therefore, dV / V =
E
2 ×109 ( Pa)
by 3.92%.
To get to energy, we consider the work of compression:
dU = − PdV
where dU is a change in recoverable internal energy U, P is the pressure applied to water, and dV
is change in volume corresponding to change in internal energy. Therefore,
g=9.81 m/s2, ρ=1000 kg/m3
dV = −
dU
=
P
1 . 4 × 10 9 ( N .m )
= 17 . 84 m 3
⎛ kg ⎞
⎛ m ⎞
9 . 81 ⎜⎜ 2 ⎟⎟ × 1000 ⎜⎜ 3 ⎟⎟ × 8000 ( m )
⎝m ⎠
⎝s ⎠
Actually, 8000 m is the final depth and we should use the average pressure here, so that change in
volume would actually be about twice this, or 35.7 cubic meters.
The minus sign appears because dV is negative under compression by a positive applied pressure
which also increases the internal energy.
dV
= 0 . 0392
,
dV = 17 . 84 m 3 → V = 910 m 3 So, roughly speaking, a good part of a
V
megalitre would have to be compressed to 8 km to store 1.4 ×109 J of energy in it.
h) E = mc2, thus m = E / c 2 = 1.4 × 109 ( J ) /(3 × 108 ) 2 (m 2 / s 2 ) = 1.56 × 10−8 (kg ) (tiny!)
1.4 ×109 ( J )
= 1944.4(m 2 ) (roughly a football field for an hour)
i) A =
2
200(W / m ) × 3600( s )
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2) Consider the heat, work and energy exchanges, and the material properties, associated with
the following two events. Account for as many of the observations (or expected observations)
as you can.
(a) It is a warm, calm summer morning and you are on vacation in Ontario. You walk beside
a small, flat, shallow, calm pond of water. You pick up a warm rock from the side of the
pond, a toss it with a splash into the middle of the pond. Consider first the rock and then
the pond as your control system. Try to be as complete as possible, and consider the
signs of the various heat, work and energy terms, using the convention we presented in
class (i.e., taking heat transfer TO the control system, and work done BY the control
system, as positive). Assume the duration of this event is a minute or two.
(b) You’re visiting a Caribbean island and one morning wake up early to take walk along
ocean shore. You notice an off-shore breeze keeps the temperature feeling quite cool,
though the morning sun is warm and pleasant. By mid-morning the beach is already too
hot to walk on except at the water’s edge, the air appears calm with little breeze, and the
day promises to be scorching. Yet, by mid-afternoon an ocean (on-shore) breeze has
picked up and you notice thick clouds forming on the hills standing inland from the shore.
You are told that rain in the heights is almost a daily occurrence.
Solution: Part a
Rock as control system:
a. Sun transfers heat to the rock (+heat input, + energy exchange)
b. You do work on the rock ( - work, + kinetic energy increase for the rock) first
to lift and then to throw the rock (using stored chemical energy in your body);
almost certainly some heat transfer too from the warm rock to the relative cool
hand; as soon as you pick it up, its radiant energy flux terms will shift from its
original location and surroundings to its new one.
c. Rock does work on the air ( + work) as it flies through the air to the pond, and
dynamically changes KE to gravitational PE as it moves vertically up and
down.
d. Rock does work on the water ( + work) causing change in surface tension
energy and displacing some of the surface which creates a splash (which
creates a 2nd order set of waves and impacts), surface waves which propagate
and gradually decay into heat, and acoustic energy which spreads out from the
impact location as a sequence of sound waves
e. Rock transfers heat to the pond (-heat input, - energy exchange)
f. Rock loses gravitational potential energy, as it sinks in the pond ( - PE, + KE)
until it hits the bottom of the pond and the rock does work on the sand at the
bottom moving it out of the way (+work). Though all its motion fluid motion
and turbulent eddies are formed (translational and rotational KE) which
gradually decay to thermal form.
Pond as control system:
− Sun transfers heat to pond (+ heat input, + energy exchange) – some if
reflected in a complex light-water interaction process that depends on the
angle and frequency of the light and state of the water surface
− Some photons will transmit through the pond water to strike the bottom of the
bond, where they will be absorbed and cause warming
− Some photons will likely be used by plants and algae to complete
4
−
−
−
−
photosynthesis and the creation of organic high energy molecules like sugars,
starches and cellulose
Air probably transfers heat to pond (+ heat input, + energy exchange)
Rock does work on the water by moving it through the application of force (work, + KE)
Almost certainly some wind-water coupling could occur that might generate
some waves, the energy of which might part build up (accumulate) and partly
decay
There will likely be an evaporative exchange, which will transfer some heat
energy from the water and (eventually) to the atmosphere (at time of
condensation)
Question 2, Part b
The offshore breeze in the morning is because, at night, the land is cooler than the water.
This is due to the three factors we discussed in class: photons penetrate more deeply in
water, being spread over a larger mass; the water has a higher heat capacity, so the
temperature raises less for a given unit of energy input; and evaporative cooling is
invariably more efficient over open water than on the land.
In any case, the cold air will then spread out under and displace the warmer air above the
ocean; its buoyancy will cause it to sink and spread through the relatively water area of
the ocean. Through the day, the air above the land increases in temperature with
exposure to the Sun until eventually there is little temperature difference between the
land and water; consequently, the breeze becomes gradually weaker and then reverses.
The land itself does not absorb a lot of heat energy so the very top surface of the land will
be hot, but it does not penetrate down much further. A lot of the light is reflected which
will quickly heat the surface air. The ocean, on the other hand, will absorb great amounts
of thermal energy. Because of this, the air directly above the water will not get as hot.
However, because of the 'thermal inertia' of the body of water, the ocean will hold more
heat than the land. This heat can be slowly dispersed through out the night, leading to a
more steady temperature through out the daily cycle. In mid-afternoon, the temperature
difference between the land and water reaches its maximum, at which point the cooler air
above the ocean travels inland displacing the warm air. As this moist air travels up the
hill, it reaches an altitude where it condenses and might even become unstable, possibly
producing a (sometime heavy) fall of rain.
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