PART 3
THERMODYNAMICS
CHAPTER 19 TEMPERATURE AND HEAT
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Problem
3. Normal room temperature is 68OF. What is this in
Celsius?
Solution
Equation 19-3, solved for the Celsius temperature,
gives Tc = i(TF - 32) = 5(68 - 32)/9 = 20°C.
Sections 19-4 and 19-5: Temperature and
Heat, Heat Capacity and Specific Heat
Problem
14. The average human diet contains 2000 kcal per
day. If all this food energy is released rather than
being stored as fat, what is the approximate
average power output of the human body?
Solution
pa, = ( 2 x 1 0 ~cal/d)(4.184 J/cal)(l d/86,400 S) =
96.9 W.
Problem
19. A circular lake 1.0 krn in diameter averages 10 m
deep. Solar energy is incident on the lake a t an
average rate of 200 W/m2. If the lake absorbs all
this energy and does not exchange heat with its
surroundings, how long will it take to warm from
10°C to 20°C?
Solution
Since the energy absorbed by ttw lake equals the solar
power times the time, At = A Q / P = ,CAT+
(200 WIm2)A, where m/A is the mass per unit area of
iake sukace: Therefore: .
A t = (lo3 kg/m3)(10 m)(4184 J/kg-K)(10 K)/(200 w/m2)
= 2.09~10' s = 24.2 d.
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Problem
27. A stove burner supplies heat at the rate of 1.0 kW,
a microwave oven at 625 W. You can heat water in
the microwave in a paper cup of negligible heat
capacity, but the stove requires a pan whose heat
capacity is 1.4 kJ/K. (a) How much water do you
need before it becomes quicker to heat on the
stovetop? (b) What will be the rate at which the
temperzture of this much water rises?
Solution
The temperature rise per second is equal to the heat
supplied per second (i.e., the power supplied if there
are no losses) divided by the total heat capacity of the
water and its container: AT/At = (AQ/At)/Cto, =
P/Ctot (Equation 19-4 divided by At). The total heat
capacity is Ctot= CW Cent, provided the water and
container both have the same instantaneous
temperature. (This assumes that heat is supplied
sufficiently slowly that the water and container share
it and stay in instantaneous thermal equilibrium.) For
the paper cup used in the microwave oven, C,,, x 0,
whereas for the pan used on the stove burner, Cent =
1.4 kJ/K. Thus, the rate of temperature rise is
625 W/Cw for the microwave and 1.0 kW/(Cw
1.4 kJ/K) for the stove burner. (a) When CW=
mwcw is small, the microwave is faster, whereas
when Cw is large, the stove burner is faster. (To see
this, plot both rates as a function of Cw.)The rates
of temperature rise are equal for Cw = mw x
(4.184 kJ1kg.K) = (1.4 kJ/K)(0.625)/(1 - 0.625) =
2.33 kJ/K. Therefore, mw = (2.33 kJ/K)+
(4.184 kJ/kg-K) = 0.558 kg. (b) For mw in part (a),
the rate of temperature rise is ATlAt = (0.625 k W ) s
(2.33 kJ/K) = 1.0 kWl(2.33 1.4)(kJ/K) = 0.268 K/s.
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-Problem
35. A piece of copper at 300°C is dropped into 1.0 kg
of water at 20°C. If the equilibrium temperature is
25OC, what is the mass of the copper?
Solution
Let us assume that all the heat lost by the copper is
gained by the water, with no heat transfer to the
container or its surroundings. Then -AQcu = AQw
(as in Example 19-3) or -mcuccu(T - T'u) =
mwcw(T - T w ) Solving for mcu (in terms of the
other quantities given in the problem or in Table 1 9 1 )
one finds mcu = mwcw(T - Tw)/ccu(TcU- T)=
(1 kg)(4184 J/kg-K)(25 - 20) K/(386 J/kg.K) x
(300 - 25) K = 0.197 kg.
Problem
39. The top of a steel wood stove measures
90 cmx40 cm, and is 0.45 cm thick. The fire
maintains the inside surface of the stove top at
310°C, while the outside surface is a t 295OC. Find
the rate of heat conduction through the stove top.
Solution
Assuming a steady flow of heat through the
90x40 cm2 = 0.36 m2 area face, with no flow through
the edges, we can use Equation 19-7 and Table 19-2:
H = - k A A T / A x = -(46 W/m.K)(0.36 m2)(295"C 310°C)/(0.45 cm) = 55.2 kW. (The heat flow is
positive, for x going from the inside of the stove to the
outside, because the temperature gradient, ATIAx, is
negative.)
Problem
47. (a) What is the R-factor for a wall consisting
of +-in. pine paneling, 72-11 fiberglass insulation,
:-in. pine sheathing, and 2.0-mm aluminum
siding? (b) What is the heat-loss rate through a
20 ftx8 ft section of wall when the temperature
difference across the wall is 55OF?
Solution
(a) The wall consists of conducting slabs of different
materials and thickness, of the same area (connected
in "series"), so the discussion following Equation 19-10
shows that the R-factor of the combination is the sum
of the individual R-factors, 77,= spine Qberglass
RA1.
The first and last terms we calculate from
Equation 19-11; the middle term is given:
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(We remembered to express A x in inches and k in
Btu-in-/h.ft2.F0.T h e contribution of the aluminum
siding is negligible.) (b) The thermal resistance of a
20x8 ft2 area of such a wall is R = (12.3120 x 8) x
(h.FO/Btu) (see solution to Problem 43), so the
heat-flow through the wall, for a steady temperature
- Tinside= A T ) , is
difference of -55 F0 (i.e., Toutside
H = -AT/R = (55 F0)(20 x 8/12.3)(Btu/h-FO) =
715 Btu/h.
Problem 47 Solution.
Problem
55. The average human body produces heat a t the
rate of 100 W and has total surface area of about
1.5 m2. What is the coldest outdoor temperature
in which a down sleeping bag with 4.0-cm loft
(thickness) can be used without t h e bodv
temperature dropping below 37OC:' Consider only
conductive heat loss.
Solution
Assume that in thermal energy balance, the rate of
heat generation by a human body equals the rate of
heat-flow lost by conduction through the sleeping bag,
i.e., 100 W = -kA AT/Ax. T h e sleeping bag may be
considered t o be closefitting (same surface area as the
body), consisting of goose down insulation (see
Table 19-2) and a negligible fabric shell. Then
100 W = -(0.043 iV/m.K)(1.5 m2) x ( T - 3T°C)+
(0.04 m), or T = 37OC - 62.0°C = -25OC = -13OF.
Problem
57. Scientists worry that a nuclear war could inject
enough dust into the upper atmosphere to reduce
significantly the amount of solar energy reaching
Earth's surface. If an 8% reduction in solar input
occurred, what would happen to Earth's 287-K
average temperature?
Solution
If we assume that the Earth's average temperature is
proportional to the one-fourth power of the effective
solar intensity (Tav S1I4),as explained in the text's
application to the greenhouse effect and global
warming, then reducing the intensity t o 0.925' alters
the average temperature according to Tlv = Tavx
(0.92)'/" This would result in a decrease in the
present Tav = 287 K of ATav = Tav- Ti, =
[I - (0.92)'/'](287 K) = 5.92 K.
CHAPTER 20
THE THERMAL BEHAVIOR OF
MATTER
ActivPhysics can help with these problems:
Activities 8.1-8.4
Section 20-1:
Gases
Problem
1. At Mars's surface, the planet's atmosphere has a
pressure only 0.0070 times that of Earth, and an
average temperature of 218 K. What is the volume
of 1 mole of the Martian atmosphere?
Solution
The molar volume of an ideal gas a t S T P for the
surface of Mars can be calculated as in Example 20-1.
However, expressing the ideal gas law for 1 mole of
gas a t the surfaces of Mars and Earth as a ratio,
PMVhf/TbI = PEVE/TE, and using the previous
numerical result, we find Vh1 = ( P E / P ~ [ ) ( T . ~ [ / T E ) X
VE = (1/0.0070)(218/273)(22.4x10-3 m3) = 2.56 m3.
Problem
5. If 2.0 mol of an ideal gas are a t an initial
temperature of 250 K and pressure of 1.5 atm,
(a) what is the gas volume? (b) The pressure is now
increased to 4.0 a t m , and the gas volume drops to
half its initial value. What is the new temperature?
(b) The ideal gas law in ratio form (for a fixed
quantity of gas, Nl = N2) gives:
Tz
=
(----)
Solution
Use the ideal gas law (Equation 20-1) in ratio form, to
compare two different states: P, Vl/P2V2 =
N1Tl/N2T2. Since the balloon contains the same
number of molecules of gas (if none escape), N1 = N2,
and V2 = ( P ~ / P Z ) ( T ~ / T =
~ )(1/0.65)(263/293)
V~
x
(8.0 L) = 11.0 L. (Note that absolute temperatures
must be used, and that any consistent units for the
ratio of the other quantities conveniently cancel.)
Problem
15. In which gas are the molecules moving faster:
hydrogen (H2) a t 75 K or sulfur dioxide (SO2) at
350 K?
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Solution
Comparing the thermal speeds, uth =
for
Hz (mass 2 u) and SO2 (mass 2 64 u) a t the given
temperatures, we find uth(H2)/uth(S02) =
~ ( T HITSO,
,
)(mso, l m ~ , =
) J(75/350)(64/2) = 2.62;
the hydrogen is faster.
Solution
(a) Frorr~Equation20-2:
4.0 atm
1.5 atm
Problem
9. A helium balloon occupies 8.0 L a t 20°C and
1.0 atrn pressure. The balloon rises to an altitude
where air pressure is 0.65 atm and the temperature
is -10°C. What is its volume when it reaches
equilibrium a t the new altitude? (Neglect tension
forces in the material of the balloon.)
(F)
(250 K) = 333 K.
Jm,
Problem
30. How much ice can a 625-W microwave oven melt
in 1.0 min, if the ice is initially a t O°C?
Solution
m = Q/Lf
112 g.
= Pt/Lf =
(625 W)(60 s)/(334 kJ/kg) =
Problem
37. A 100-g block of ice, initially at -20°C, is placed
in a 500-W microwave oven. (a) How long must
the oven be on to produce water at 50°C?
(b) Make a graph showing temperature versus
time during this entire interval.
Solution
(a) To bring the ice to O°C requires heat: Q1 =
mci,,AT = (0.1 kg)(2.05 kJ/kg.K)[O0C - (-20°C)] =
4.10 kJ. To melt the ice requires Q2 = mLf =
(0.1 kg)(334 kJ/kg) = 33.4 kJ. Finally, to bring the
meltwater from 0°C to 50°C requires heat Q3 =
mcWat,,AT = (0.1 kg)(4.184 kJ/kg.K)(50°C - 0°C) =
20.9 kJ. Power must be supplied for a time t =
Q t o t / P , or t = (4.10 + 33.4 20.9) kJ10.5 kW = 117 s.
(b) It takes times t l = Q1/0.5 kW = 8.2 s, t z = 66.8 s,
and t 3 = 41.8 s for the preceding steps, respectively.
Since the power input is constant, the temperature is
linear for steps 1 and 3, and, of course, constant
during melting, as shown.
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Problem 37 Solution.
Problem
47. A bowl contains 16 kg of punch (essentially water)
a t a warm 25OC. What is the minimum amount of
ice a t O°C that will cool the punch to O°C?
Solution
Assume t h a t the only heat transfer is between the
punch and the ice. To cool to O°C, A Q =
(16 kg)(4.184 kJ/kg.K)(25OC - 0°C) = 1.67 MJ of
heat must be extracted from the punch. A minimum
mass m = AQ/Lf = 1.67 MJ/(334 kJ/kg) = 5.01 kg
of ice at O°C could do this, but the punch would be
diluted with 5.01 kg of melt-water. (To reduce the
dilution, sufficient ice at a temperature below O°C is
needed.)