Re: Speed at Which Point Charge in Approaching origin / loss of charge at constant ra
Thank you Simon -- I'll give this a go...
".. I don't see how that is "introductory" ... what level is this to be done at?" = My thoughts exactly...
...nd day of Principles of Physics II...
Re: Speed at Which Point Charge in Approaching origin / loss of charge at constant ra
My profound apologies = Yes, in cm.
Can you please help a little more...an equation?...Yes, I am really lost on b
Re: Speed at Which Point Charge in Approaching origin / loss of charge at constant ra
1.71 kg is correct.
Still lost on b; F = 16.794 N and it is only in y direction
a)Point charge Q = +7.25 μC is fixed at the origin. Point charge q = +1.60 μC is now carefully placed on the positive y-axis, and it floats at (0,7.88). Find the mass of q.
b) Refer to question a. Suppose that, as charge q is floating, the point charge Q at the origin begins to lose its...
Pressure, Volume and Temp Change...WORK done???
n = 9.95 moles of an ideal gas are slowly heated from initial pressure P1 = .922 atm and initial volume V1 = 679 L to final pressure Pf = 1.16 atm and final volume Vf = 1073 L. Find T1, the initial temperature of the gas. If the plot of this...
Re: Thermodynamics problem -- need to find amount of water that boils
Yes -- kilograms, so that is not the issue. Specific question is: Find the amount of water that boils. But below question it also states "Note: reaching the boiling point is not enough, the question asks for the amount of...
Re: Thermodynamics problem -- need to find amount of water that boils
wait wait wait...
If we start with .658 kg of water and .148 vaporizes...what is the amount of water that boils?
Re: Thermodynamics problem -- need to find amount of water that boils
Are we sure there is a phase change for the lead? If so, is it definitely the full amount?
Re: Thermodynamics problem -- need to find amount of water that boils
FYI: Below answer is incorrect. Please help!! What am I doing wrong now???
257000 + 303362.8 = 223931.7444 + Q
Q = 336.4310556 kJ
m = 336.4310556 / 2260 kJ/kg = .148 kg
Re: Thermodynamics problem -- need to find amount of water that boils
Ok...what about this:
257000 + 303362.8 = 223931.7444 + Q
Q = 336.4310556 kJ
m = 336.4310556 / 2260 kJ/kg = .148 kg
Re: Thermodynamics problem -- need to find amount of water that boils
.658 Kg is the mass of the water; it's given in the problem.
How about this:
Energy released when lead turns from liquid to solid = (10.28)(25kJ/Kg) = 257,000J
257000 + (10.28)(130)(327 - T) = (.658)(4186)(T - 18.7)
T =...
Re: Thermodynamics problem -- need to find amount of water that boils
Evidently not. One last chance to get this right...little help please on needed equation.
Re: Thermodynamics problem -- need to find amount of water that boils
Is the specific heat of lead needed to calculate the heat loss of the lead in addition to the energy released from the phase change (change from liquid to solid)?
Re: Thermodynamics problem -- need to find amount of water that boils
What is the equation i need to set up?????
There is energy released when lead turns from liquid to solid, there is the heat needed to raise water to 100 degrees celsius, there is heat needed to change water to steam, there...
Re: Thermodynamics problem -- need to find amount of water that boils
Both...but I guess that is not right. Final temp for everything that remains is 100?
Re: Thermodynamics problem -- need to find amount of water that boils
ok....
Energy released when lead turns from liquid to solid at melting point:
(10.28 kg) x (25kJ/Kg) = 257kJ
Heat to raise water to 100 degrees Celsius:
(.658 kg) x (4186) x(100-18.7) = 223.9317kJ
Heat to change water to...
n = 9.95 moles of ideal gas are slowly heated from initial pressure Po = 93398.6 Pa and initial volume Vo = .679 m^3 to final pressure Pf = 117508 Pa and final volume Vf = 1.073 m^3. If the plot of this process on a P-V diagram is a straight line, find W, the work done on the system:
If the...
Re: Thermodynamics problem -- need to find amount of water that boils
Ok...um...how is this Wrong??
Heat loss by lead = (10.28)(130)(327.5-100) = 304031 J
Heat gain by water = (.658)(4186)(100-18.7) = 223931.7444 J
Mass of water turned to steam = Q / L = (304031 - 223931.7444) = 80099.25556...
Re: Thermodynamics problem -- need to find amount of water that boils
Heat required to change the phase of the water is the Latent heat...right? Equation is Q = mL where L = 2260 kJ/kg. Right?
Thermodynamics problem -- need to find amount of water that boils
Suppose molten (liquid) lead, mass = 10.28 kg, is at its melting point. The lead is poured into water of mass = 658 g and initial temperature T = 18.7 degrees C. Find the amount of the water that boils. Assume no heat loss.
No...