course Phy232
I completed the problems in this assignment but I am pretty sure that I didn't get hardly any of them wrong. This is one of the problems I have. I have never been able to fully understand some parts of Physics but I can pick out equations and I think I got the equations right but not sure about anything else. Thank You!
76: H=AeoT^4 (heat current in radiation
A=4Pir^2
=4Pi(1.50)^2
=28.27
H=(28.27)(0.35)(5.67(40)x10^-8)(3000)^4
H=9.8945(2.268x10^-6)(8.1x10^13)
H=1.8178x10^-9
None of the above quantities include units. So what I can tell by looking at your work is the following:
You have calculated the radiative thermal current from the surface a sphere of radius 1.50 (units unspecified) with emissivity 0.35 and surface temperature 3000 (units unspecified). There's a number 40 in the middle of the Stephan-Boltzmann constant that doesn't belong there, and the units of the constant are not specified (though this is the value of the constant in watts / (m^2 Kelvin^4).
It looks like most of what you're doing is correct, but you need to tell me what you're trying to find and how it connects to the information you're given, and you need to include the units.
81.
(A): To+T density=p=po(1-Beta delta T)
L=Lo(1+alpha delta T)
B varies with the temperature and it is valid only for small temperatures because it is a solid.
I assume by B you mean Beta, and that you are using p to stand for the Greek letter rho (if so, better to write rho, but I think I can tell what you mean)?
(B): Copper cube=side lenths=1.25cm at 20.0 degrees Celcius
Find change in its volume and density when temp is increased to 70.
V=L^3=(1.25^3)=1.95
Delta T: 70.0-20.0=50 degrees celcius
Delta L= alpha L1 Delta T= (17x10-6(C)-1)(1.25)(50)
=1.06x10-4
Delta V=3alpha delta T+3(alpha delta T)^2+(alpha delta T)^3
=3(17x10^-6)(50)+3((17x10^-6)(50))^2+((17x10^-6(50))^3
=0.00255+2.1675x-6+6.14125x10^-10
= 2.55x10^-3
L = L0 ( 1 + alpha `dT) so
V = L^3 = L0^3 ( 1 + 3 alpha `dT + 3 (alpha `dT)^2 + (alpha `dT)^3). Since alpha `dT is small, (alpha `dt) ^2 and alpha^3 are so small they're negligible, so we have
V = L0^3 ( 1 + 3 alpha) = V) ( 1 + 3 alpha).
So beta = 3 alpha.
Not sure what the stated goal of the problem is, but you seem to have most of it.
92.
(A): Well stree is a quantity that is proportional to the force causing a deformation. Bulk modulus which measures the resistance of solids or liquids to change is their volume. If it is not allowed to expand then they will work together to make sure none expand.
unclear what 'they' refers to, but this is the beginning of a good statement
(B): Tk=Tc+273.15
= 15+273.15=288.15
= -(70x10^11Pa)(1.2x10^-5)(288.15)
= -2/42x10^10
You state, in part, that
288.15
= -(70x10^11Pa)(1.2x10^-5)(288.15)
which is not a true statement. You intended train of thought, not equality, but you didn't say what your train of thought was. Presumably your second step appears to include the pressure and some coefficient of expansion (hopfully, given the context, the coefficient of volume expansion); however the quantity that appears to be pressure might also be bulk modulus.
Again if you're off, you're not far off. Can you clarify a little more?
97:
(A): dQ= 4-10= 30K
= nCdT
30=1.50(186405.52)(40)
= 2.6x10^7
30=1.50(186405.52)(40) just isn't so; the right- and left-hand sides aren't equal.
The context is clear but I can't tell what 186495.52 means or what it's doing there.
Should be easy for you to clarify.
100) c=29.5J/molxK + (8.20x10^-3J/molxK^2)T
Looks like the specific heat of a non-ideal diatomic gas at constant pressure
dQ=3molxCxdT=3mol (29.5+(8.2x10^-3)T)xdT
F(t)=29.5xT+(8.2+10^-3)x T2/2
it appears that you've correctly found an antiderivative, which should lead you to a correct quantity of thermal energy
(6696.5+8.2x10^-3)x364.5
unclear where some of these numbers come from, what their units are, what they mean
103:
(A): C1=4.19
dT1=Tf-T1 and DT2=Tf-T2
C2M2(Tf-T2)+c1m1(Tf-T1)=0
(4.19)(0.0350)(Tf-100)+(4.19)(0.0950)(Tf-0)=0
0.14665(Tf-100)+0.39805(Tf-0)=0
0.14665tf-14.665+0.39805Tf=0
0.14665tf+0.39806Tf=14.665
Tf= 26.92 degrees C
this appears to be some sort of calorimetry problem; with temperatures of 100 and 0 it would appear that ice and steam might be involved
however heats of fusion and vaporization are not considered; it seems likely they would be involved
(B): No ice and NO steam
106:
Q=mcdelta T
=(0.525)(420)(56)
= 220.5(56)
= 1.2x10^4J
the situation and units here are unclear; I suspect the problem is more complex than this simple relationship would indicate
109:
(A): Rate of heat flow through the door
H=dQ/dT=KA(Th-Tc/L)
=(0.120)(9.f)(20-(-8)/1.8)
= (1.14)(15.56)
I suspect that your calculation 20-(-8)/1.8 was done incorrectly; that looks like the temperature gradient, which should be (change in temperature) / (change in position), in this case numerically equal to 12 / 1.8.
Never mind. You multiplied by that 9.f ('f' obviously a typo, but the quantity apparently representing the area of the door). Now the 15.56 makes sense, except for its lack of units.
You then multiply by thermal conductivity (no units specified), so your result makes sense (17.75, not 17/74; obvious typo, no problem there).
= 17/74
(B): H=(0.80)(12)(20+8/12.45)
= 1.6(2.25)
= 3.6
17.74+3.6= 21.34
very similar calculation but unsure what the goal of the problem is
112:
(A):Temperature of Junction Point?
Hcopper=KcA(100C-T)/Lc
Hsteel= KsA(100C-T)/Ls
Hbrass= KbA(100C-t)/Lb
not sure whether this is a parallel or series junction; if parallel then the common end temperatures 100 C and T make sense; if not you need unknowns for the intermediate temperatures, and you'll get some simultaneous equations
=(385)(100-T)/0.130=(50.2)(T-0)/0.240=(109)(T-0)/0.18
=38500-385T/0.130=50.2/0.240=(109)(T-0)/0.180
=38500-385T/0.130=50.2T+385T-109.0T/0.240=0.180
= 29615.4=326.2T=0.180
=29615.22/326.2=91 degrees C
(B)Heat Current in each rod?
Hc=(385)(100-91)/0.130= 3465/0.130= 26,652.8
Hs=(50.2)(91-0)/(0.240)=4568.2/0.240=19034.2
Hb=(109)(91-0)/0.18= 9919/0.180= 55,105.6"
You have the right relationships here, but the setup and the goal of the problem is unclear. If you can clarify I can comment further.
Your work looks pretty good. It certainly appears that you're up to the level of this course, and you're communicating a lot of good information.
I don't carry a copy of the text around, and unfortunately don't have electronic copies of the problem statements (which would be easier for both of us). So I've reverse-engineered your solutions as best I could from the information you've provided.
This should give you a good idea of what I need to see when you submit work of this nature. You're telling me 90% of what I need to know, and it should be easy to insert the rest.
I'll look over the other questions you've sent, but won't have time to reverse-engineer all of them (the 20 minutes I spent doing so on this assignment is the most fun I've had today, and I mean that seriously, but other students are waiting), so when they're posted at your access page later this evening it will usually be with a note asking you to insert a little more information and resubmit. I will however comment if I see any place where you clearly seem to be off track.