Download Report

Problem 1
What is the molar mass of glucose, C6H12O6?
a. 180 g/mol
c. 160 g/mol
b. 144 g/mol
d. 360 g/mol
Link within Knovel
Solution
Included with this link is the hint “Try ‘Tools’ at the top pf the page”. The link brings the player to the
Knovel homepage and in the Tools dropdown at the top, one may find a link to the Periodic Table. It
shows that the atomic mass of a carbon atom is 12.01 atomic mass units (u), so the molar mass is 12.01
grams per mole. There are 6 carbon atoms in 1 molecule of glucose, so carbon contributes
6*12.01=72.06 u, or 72.06 grams per mole. Performing this for all of the elements that comprise the
molecule:
Carbon:
6*12.01 g/mol
Hydrogen:
12*1.008 g/mol
Oxygen:
6*16.00 g/mol
180.156 g/mol
Problem 2
What is the molar mass of water, H2O?
a. 33 g/mol
b. 44 /mol
b. 18 g/mol
d. 50 g/mol
Link within Knovel
Solution
This link brings the player directly to the Knovel Periodic Table. Using the atomic masses of hydrogen
and oxygen provided, the molar mass of water is found to be just greater than 18 g/mol:
Hydrogen:
2*1.008 g/mol
Oxygen:
1*16.00 g/mol
18.016 g/mol
Problem 3
6CO2 + 6H2O (+light)  C6H12O6 + 6O2
Given this balanced chemical equation describing photosynthesis, and an excess of CO2, how much
water is needed to produce 100 grams of glucose?
a. 10 g
b. 60 g
c. 3.33 g
d. 10.8 kg
Link within Knovel
Solution
This link brings the player to a page of a resource with an example that is almost identical to the
problem at hand. The pertinent pieces of information to extract from the provided chemical equation
are the coefficients of water and glucose, 6 and 1 respectively. These coefficients tell the reader that to
produce 1 mole of glucose 6 moles of water are needed (provided there is enough CO2 to complete the
reaction). Now following the example provided in the text and using the molar masses calculated in
problems 1 and 2:
𝑚𝑜𝑙 𝑜𝑓 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 6 𝑚𝑜𝑙 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
18 𝑔
100 𝑔 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 ∗
∗
∗
= 60 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
180 𝑔
1 𝑚𝑜𝑙 𝑜𝑓 𝑔𝑙𝑢𝑐𝑜𝑠𝑒 1 𝑚𝑜𝑙 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
Problem 4
The Maclaurin series of sin(x2) expanded to four terms is
𝑥6
𝑥 10
𝑥 14
−
+
3!
5!
7!
𝑥3
𝑥5
𝑥7
− 3! + 5! − 7!
𝒙𝟔
𝒙𝟏𝟎
𝒙𝟏𝟒
−
𝟓!
𝟕!
𝑥5
𝑥7
𝑥9
2
𝑥 − 3! + 5! − 7!
a. −𝑥 2 +
b. 𝒙𝟐 − 𝟑! +
c. 𝑥
d.
Link within Knovel
Solution
This hint brings the player to a page of a resource that provides Maclaurin series of some key functions
expanded to a few terms. The expansion of sin(x) is provided to four terms as
sin(𝑥) = 𝑥 −
𝑥3 𝑥5 𝑥7
+ −
3! 5! 7!
and the domain is all x. Replacing x above with x2, and making use of the equality (xa)b=xab, one obtains
the expression provided in choice b.
Problem 5
The sum of the first five terms of the Maclaurin expansion of e2 is
a. 7.389
b. 7
c. 5
d. 6.333
Link within Knovel
Solution
This hint brings the player to the same page as the previous one. It provides the Maclaurin expansion of
ex to five terms as:
𝑥2 𝑥3 𝑥4
𝑒𝑥 = 1 + 𝑥 + + + + ⋯
2! 3! 4!
for all x. In this particular problem, x=2. Substituting in the value of x and evaluating the factorials in the
denominators, one obtains:
4 8 16
𝑒2 = 1 + 2 + + +
+⋯
2 6 24
𝑒2 = 7 + ⋯
The sum of the first five terms is simply 7.
Problem 6
The difference between the true value of e3, and the sum of the first five terms of the Maclaurin
expansion is approximately
a. 0
b. 3.71
c. 2.718
d. 𝜋
Link within Knovel
Solution
The true value of e3 is approximately 20.0855. The link provided brings the player to the same page that
the hint in the previous two problems did. It shows that the expansion of ex, to five terms, is
1+𝑥+
𝑥2 𝑥3 𝑥4
+ + +⋯
2! 3! 4!
Substituting in x=3 and evaluating, one obtains:
9 27 81
1+3+ +
+
+⋯
2 6 24
16.375 + ⋯
So the difference between the true value and the sum of the first five terms of the expansion is
approximately 3.71
𝑒 3 − 16.375 ≈ 3.71
Included with problems 7 through 9 are the instructions: “Consider the function f(x) = -2x6+x5-x4+x3+5x2 ”
Problem 7
What are the roots?
a. -1, 0, 1.3711
c. 1, 0, 1.3711, 2.7422, 5, 7.142
b. 0, 1.3711, 2.7422
d. -1, 1, 1.3711, 5, 7.142, 8
Watch the Functions Palette (Part II) How-to video here:
Link within Knovel
The link included with this problem brings the player to a video that will walk one through a process to
answer questions 7 through 9. A problem similar to this one is considered at approximately 3:51 in the
video. In the Knovel Interactive Equations homepage, click the blue button on the right to open a new
interactive worksheet. Open it in its own tab if desired. Click anywhere in the whitespace then begin
typing what is italicized here.
f(x [space] [colon] -2x^6 [space] + x^5 [space] –x^4 [space] + x^3 [space] + 5x^2 [enter]
This will define the function so that it may be called. Notice that a second parenthesis is not typed.
In another whitespace area type
p [colon] solve (f(x [space] [comma] x [enter]
p=
Knovel will return a vector containing the three roots,
−1
p=( 0 )
1.3711
Problem 8
What is the area below the curve, for which f(x)>0?
a. 1.1173
c. 9.81
b. 6.67
d. 3.4793
Link within Knovel
Solution
This hint is identical to the last one, and the problem is referring to the same function used in problem 7,
which is also defined in the instructions provided with this problem. The solution to a problem very
similar to this one is addressed in the video immediately following the portion that addressed the
subject of the previous problem. Continue working in the same interactive worksheet.
With p defined as the vector with the three roots of the function, in any whitespace type
a [colon] int(f(x [space] [comma] x [comma] p [left bracket] 1 [space] [comma] p [left bracket] 3 [space]
[space] =
Notice how Knovel interprets the keystrokes, as indicated by the emerging integrand and movement of
the cursor. Knovel will return 3.4793.
To see what this looks like click any whitespace and type @. In the area under the graph that appears
type f(x [space] [enter]. Knovel will produce a graph that shows the function. In more familiar symbols,
Knovel just evaluated the following definite integral:
1.3711
𝑎= ∫
𝑓(𝑥) 𝑑𝑥
−1
Alternatively, one could have used Knovel to compute the sum of each of the areas on either side of the
y axis:
0
1.3711
𝑎 = ∫ 𝑓(𝑥)𝑑𝑥 + ∫
−1
𝑓(𝑥) 𝑑𝑥
0
but the former approach is quicker and just as accurate.
Problem 9
What is the slope of the tangent at x=6?
a. 0
c. -87,528
b. 87,528
d. 14
Link within Knovel
Solution
This hint is identical to the previous two. The video begins a problem nearly identical to this one at
approximately 0:33. Continue working with the same interactive worksheet that has already been
created. Click in any whitespace and type
s(x [space] [colon] [derivative] f(x [right arrow] [right arrow] x [right arrow] [enter]
In place of [derivative] click the derivative symbol in the functions palette on the right.
s(6 [space] =
Knovel will return -87,528.
Included with problems 10 through 13 is the following introduction:
An uninsulated pipe, with an outer diameter of 6 inches, and a surface temperature of 300°F passes
through a room in which the air and the walls are at 77°F. The emissivity of the pipe is 0.7, and the heat
𝐵𝑇𝑈
transfer coefficient associated with this flow is 3ℎ𝑟 𝑓𝑡 2 °𝑅𝑎 .
Problem 10
Does the pipe behave as a blackbody?
a. yes
c. impossible to determine
b. no
d. only in this environment
Link within Knovel
Solution
This hint brings the player to a list of Interactive Equations on thermodynamics. From this list, one can
search for “black”. This will return two equations. Upon examining the second, Non-Blackbody Emissive
Power one will read in the description “If a surface is not black, the emissive power of the surface
depends on its emissivity which is a surface characteristic whose value ranges from 0 to 1, with the
emissivity of a blackbody being 1.” A value of 0.8 implies not a blackbody, and such is called a gray body.
A blackbody is an idealized body that radiates energy at the maximum possible rate for a given
temperature and absorbs all radiation that falls upon it. It is used an ideal approximation for many real
scenarios.
Problem 11
What is the emissive power of the surface?
a. 99.4 BTU/(hr*ft2)
c. 571 BTU/(hr*ft2)
b. 400 BTU/(hr*ft2)
d. 2.71828 BTU/(hr*ft2)
Link within Knovel
This hint brings the player directly to the Knovel Interactive Equation Non-Blackbody Emissive Power,
the same one that was used to answer the previous problem. Open the equation worksheet. Enter an
emissivity (𝜀) of 0.7 and press enter. The Stefan-Boltzmann constant is constant, so do not change its
value or units. Enter a surface temperature of 300°F and allow the interactive worksheet to convert to
°Ra(not shown, only reflected in the answer), or use the Knovel Unit Converter to convert. (To convert
from °F to °Ra, simply add 459.7, or 460) To enter the temperature in °F, dele the F and type Ra. Tap
enter on the keyboard. Knovel will return
𝐵𝑇𝑈 1
𝐸 = 399.12
·
ℎ𝑟 𝑓𝑡 2
The most reasonable answer of those provided is choice c.
Problem 12
What is the rate of energy flow, per unit length, from the surface of the pipe to the air?
a. 131.36 BTU/hr
b. 334.5 BTU/hr
c. 1050.86 BTU/hr
d. 78.05 BTU/hr
Link within Knovel
Solution
This hint brings the player to Knovel’s Interactive Equation Convection. From the description, convection
is the method of energy transfer “between a solid surface and a moving fluid”. Open the interactive
worksheet. Replace the default value of the heat transfer coefficient, h, with 3 as given in the
introduction; the units are the same. Tap enter or return on the keyboard. The surface area of the pipe
with circular cross section and open on both ends is equal to the circumference of the cross section
multiplied by the length of the pipe. Because the problem asks for the rate of energy flow per unit
length, only the circumference needs to be considered.
𝐶 = 2𝜋𝑟 = 𝜋𝐷 = 0.5 𝑓𝑡 ∗ 𝜋. The area may be entered as 0.5 𝜋 with units of ft2.
(If the length of the pipe were given, one would multiply the circumference of the pipe by it, complete
the process to obtain a value for the energy transfer, then divide this by the length of the pipe, to find
the value per unit length. In the end, the length will cancel out, so it is not included in the calculations)
The surface temperature, Ts , is 300°F. Delete the Ra and type F, or keep the units of °Ra and convert the
300°F to 759.7°Ra (simply add 459.7, or 460). The fluid temperature, 𝑇∞ , is 77°F or 536.7°Ra. Knovel
returns
qconv = 1050.86
BTU
.
hr
Problem 13
What is the rate of energy transfer, per unit length, from the surface of the pipe to the walls of the
room?
a. 15.18 BTU/hr
b. 20.71 BTU/hr
c. 470 BTU/hr
d. 1.48 BTU/hr
Link within Knovel
Solution
This hint brings the player to the Knovel Interactive Equation Net Radiant Heat Transfer. Because the
problem asks for the rate of energy transfer between two surfaces, radiation is the method to consider.
Open the interactive worksheet. The Stefan-Boltzmann constant is a fundamental physical constant, so
it is unchanged. The gray surface area is that of the pipe (per unit length), or 0.5 ∗ 𝜋 ft2. (To enter 𝜋, use
the Lowercase Greek palette on the right.) ε is the emittance, or emissivity, of the gray surface and is
given in the introductory statement as 0.7. T1 is the temperature of the pipe, 300°F or 759.7°F. T2 is the
temperature of the walls, 77°F or 536.7°F. Knovel will return
qr = 470.84
BTU
.
hr
As mentioned in the solution to problem 10 a true blackbody does not exist, and Knovel’s reference to
T2 as the “surrounding black surface temperature” is an approximation. This simply means that the wall
does not absorb all of the energy radiated to it by the pipe; it does not affect the rate at which the pipe
radiates it.